Ruby 1.9,185个字符

Question

挑战：

构建给定文本中最常用单词的ASCII图表。

规则：

仅接受a-z和A-Z（字母字符）作为单词的一部分。
为了我们的目的，忽略套管（She == she）。
忽略以下字词（非常简单，我知道）：the, and, of, to, a, i, it, in, or, is
澄清：考虑don't：这将被视为范围a-z和A-Z中的2个不同的'字词':( don和{{1 }}）。
（可选）（现在正式更改规格为时已晚）您可能选择放弃所有单字母“字词”（这可能会使缩短忽略列表。）

解析给定的t（读取通过命令行参数指定的文件或管道输入;假设text）并构建一个具有以下特征的us-ascii：

显示22个最常用单词的图表（另请参阅下面的示例）（按降序排序）。
条形word frequency chart表示单词的出现次数（频率）（按比例）。附加一个空格并打印出来。
确保这些条形（加上空格字符空间）总是适合：width + bar + [space] + word应该是始终＆lt; = [space]个字符（确保您考虑可能不同的条形和字长：例如：第二个最常见的字可能比第一个字长很多但频率差别不大）。 在这些约束条件下最大化条宽度并适当缩放条形（根据它们所代表的频率）。

示例：

示例can be found here的文字（ Alice's Wonderland in Wonderland，作者：Lewis Carroll ）。

此特定文本将产生以下图表：

 _________________________________________________________________________
|_________________________________________________________________________| she 
|_______________________________________________________________| you 
|____________________________________________________________| said 
|____________________________________________________| alice 
|______________________________________________| was 
|__________________________________________| that 
|___________________________________| as 
|_______________________________| her 
|____________________________| with 
|____________________________| at 
|___________________________| s 
|___________________________| t 
|_________________________| on 
|_________________________| all 
|______________________| this 
|______________________| for 
|______________________| had 
|_____________________| but 
|____________________| be 
|____________________| not 
|___________________| they 
|__________________| so

供您参考：以上图表的构建频率如下：

[('she', 553), ('you', 481), ('said', 462), ('alice', 403), ('was', 358), ('that
', 330), ('as', 274), ('her', 248), ('with', 227), ('at', 227), ('s', 219), ('t'
, 218), ('on', 204), ('all', 200), ('this', 181), ('for', 179), ('had', 178), ('
but', 175), ('be', 167), ('not', 166), ('they', 155), ('so', 152)]

第二个示例（检查您是否实施了完整的规范）： 使用80替换链接的爱丽丝梦游仙境文件中you的每次出现：

 ________________________________________________________________
|________________________________________________________________| she 
|_______________________________________________________| superlongstringstring 
|_____________________________________________________| said 
|______________________________________________| alice 
|________________________________________| was 
|_____________________________________| that 
|______________________________| as 
|___________________________| her 
|_________________________| with 
|_________________________| at 
|________________________| s 
|________________________| t 
|______________________| on 
|_____________________| all 
|___________________| this 
|___________________| for 
|___________________| had 
|__________________| but 
|_________________| be 
|_________________| not 
|________________| they 
|________________| so

获胜者：

最短的解决方案（按字符计算，每种语言）。玩得开心！

编辑：表格汇总了迄今为止的结果（2012-02-15）（最初由用户Nas Banov添加）：

Language          Relaxed  Strict
=========         =======  ======
GolfScript          130     143
Perl                        185
Windows PowerShell  148     199
Mathematica                 199
Ruby                185     205
Unix Toolchain      194     228
Python              183     243
Clojure                     282
Scala                       311
Haskell                     333
Awk                         336
R                   298
Javascript          304     354
Groovy              321
Matlab                      404
C#                          422
Smalltalk           386
PHP                 450
F#                          452
TSQL                483     507

数字代表特定语言中最短解决方案的长度。 “严格”是指完全实现规范的解决方案（绘制superlongstringstring条，用|____|行关闭顶部的第一个条，考虑高频长字的可能性等）。 “放松”意味着采取一些自由来缩短解决方案。

仅包含短于500个字符的解决方案。语言列表按“严格”解决方案的长度排序。 'Unix Toolchain'用于表示使用传统* nix shell plus 混合工具（如grep，tr，sort，uniq，head，perl，awk）的各种解决方案。

Answer 1

LabVIEW 51节点，5个结构，10个图

教大象踢踏舞从来都不是很好。我会啊，跳过字符数。

程序从左向右流动：

Answer 2

Ruby 1.9,185个字符

（严重基于其他Ruby解决方案）

w=($<.read.downcase.scan(/[a-z]+/)-%w{the and of to a i it in or is}).group_by{|x|x}.map{|x,y|[-y.size,x]}.sort[0,22]
k,l=w[0]
puts [?\s+?_*m=76-l.size,w.map{|f,x|?|+?_*(f*m/k)+"| "+x}]

您可以简单地将文件名作为参数传递，而不是像其他解决方案那样使用任何命令行开关。（即ruby1.9 wordfrequency.rb Alice.txt）

由于我在这里使用字符文字，因此该解决方案仅适用于Ruby 1.9。

编辑：用换行符替换分号以获取“可读性”。：P

编辑2：Shtééf指出我忘记了尾随空间 - 修正了。

编辑3：再次删除尾随空格;）

Answer 3

GolfScript， 177 175 173 167 164 163 < / s> 144 131 130个字符

慢 - 样本文本为3分钟（130）

{32|.123%97<n@if}%]''*n%"oftoitinorisa"2/-"theandi"3/-$(1@{.3$>1{;)}if}/]2/{~~\;}$22<.0=~:2;,76\-:1'_':0*' '\@{" |"\~1*2/0*'| '@}/

说明：

{ #loop through all characters 32|. #convert to uppercase and duplicate 123%97< #determine if is a letter n@if #return either the letter or a newline }% #return an array (of ints) ]''* #convert array to a string with magic n% #split on newline, removing blanks (stack is an array of words now) "oftoitinorisa" #push this string 2/ #split into groups of two, i.e. ["of" "to" "it" "in" "or" "is" "a"] - #remove any occurrences from the text "theandi"3/-#remove "the", "and", and "i" $ #sort the array of words (1@ #takes the first word in the array, pushes a 1, reorders stack #the 1 is the current number of occurrences of the first word { #loop through the array .3$>1{;)}if#increment the count or push the next word and a 1 }/ ]2/ #gather stack into an array and split into groups of 2 {~~\;}$ #sort by the latter element - the count of occurrences of each word 22< #take the first 22 elements .0=~:2; #store the highest count ,76\-:1 #store the length of the first line '_':0*' '\@ #make the first line { #loop through each word " |"\~ #start drawing the bar 1*2/0 #divide by zero *'| '@ #finish drawing the bar }/

“正确”（希望如此）。（143）

{32|.123%97<n@if}%]''*n%"oftoitinorisa"2/-"theandi"3/-$(1@{.3$>1{;)}if}/]2/{~~\;}$22<..0=1=:^;{~76@,-^*\/}%$0=:1'_':0*' '\@{" |"\~1*^/0*'| '@}/

慢一点 - 半分钟。（162）

'"'/' ':S*n/S*'"#{%q '\+" .downcase.tr('^a-z',' ')}\""+~n%"oftoitinorisa"2/-"theandi"3/-$(1@{.3$>1{;)}if}/]2/{~~\;}$22<.0=~:2;,76\-:1'_':0*S\@{" |"\~1*2/0*'| '@}/

输出在修订日志中可见。

Answer 4

基于Transact SQL set的解决方案（SQL Server 2005） 1063 892 873 853 827 < / s> 820 783 683 647 644 630个字符

感谢Gabe提供一些有用的建议来减少字符数。

注意：添加换行符以避免滚动条只需要最后一个换行符。

DECLARE @ VARCHAR(MAX),@F REAL SELECT @=BulkColumn FROM OPENROWSET(BULK'A', SINGLE_BLOB)x;WITH N AS(SELECT 1 i,LEFT(@,1)L UNION ALL SELECT i+1,SUBSTRING (@,i+1,1)FROM N WHERE i<LEN(@))SELECT i,L,i-RANK()OVER(ORDER BY i)R INTO #D FROM N WHERE L LIKE'[A-Z]'OPTION(MAXRECURSION 0)SELECT TOP 22 W,-COUNT(*)C INTO # FROM(SELECT DISTINCT R,(SELECT''+L FROM #D WHERE R=b.R FOR XML PATH (''))W FROM #D b)t WHERE LEN(W)>1 AND W NOT IN('the','and','of','to','it', 'in','or','is')GROUP BY W ORDER BY C SELECT @F=MIN(($76-LEN(W))/-C),@=' '+ REPLICATE('_',-MIN(C)*@F)+' 'FROM # SELECT @=@+' |'+REPLICATE('_',-C*@F)+'| '+W FROM # ORDER BY C PRINT @

可读版本

DECLARE @ VARCHAR(MAX), @F REAL SELECT @=BulkColumn FROM OPENROWSET(BULK'A',SINGLE_BLOB)x; /* Loads text file from path C:\WINDOWS\system32\A */ /*Recursive common table expression to generate a table of numbers from 1 to string length (and associated characters)*/ WITH N AS (SELECT 1 i, LEFT(@,1)L UNION ALL SELECT i+1, SUBSTRING(@,i+1,1) FROM N WHERE i<LEN(@) ) SELECT i, L, i-RANK()OVER(ORDER BY i)R /*Will group characters from the same word together*/ INTO #D FROM N WHERE L LIKE'[A-Z]'OPTION(MAXRECURSION 0) /*Assuming case insensitive accent sensitive collation*/ SELECT TOP 22 W, -COUNT(*)C INTO # FROM (SELECT DISTINCT R, (SELECT ''+L FROM #D WHERE R=b.R FOR XML PATH('') )W /*Reconstitute the word from the characters*/ FROM #D b ) T WHERE LEN(W)>1 AND W NOT IN('the', 'and', 'of' , 'to' , 'it' , 'in' , 'or' , 'is') GROUP BY W ORDER BY C /*Just noticed this looks risky as it relies on the order of evaluation of the variables. I'm not sure that's guaranteed but it works on my machine :-) */ SELECT @F=MIN(($76-LEN(W))/-C), @ =' ' +REPLICATE('_',-MIN(C)*@F)+' ' FROM # SELECT @=@+' |'+REPLICATE('_',-C*@F)+'| '+W FROM # ORDER BY C PRINT @

输出

_________________________________________________________________________ |_________________________________________________________________________| she |_______________________________________________________________| You |____________________________________________________________| said |_____________________________________________________| Alice |_______________________________________________| was |___________________________________________| that |____________________________________| as |________________________________| her |_____________________________| at |_____________________________| with |__________________________| on |__________________________| all |_______________________| This |_______________________| for |_______________________| had |_______________________| but |______________________| be |_____________________| not |____________________| they |____________________| So |___________________| very |__________________| what

用长串

_______________________________________________________________ |_______________________________________________________________| she |_______________________________________________________| superlongstringstring |____________________________________________________| said |______________________________________________| Alice |________________________________________| was |_____________________________________| that |_______________________________| as |____________________________| her |_________________________| at |_________________________| with |_______________________| on |______________________| all |____________________| This |____________________| for |____________________| had |____________________| but |___________________| be |__________________| not |_________________| they |_________________| So |________________| very |________________| what

Answer 5

206

shell，grep，tr，grep，sort，uniq，sort，head，perl

~ % wc -c wfg
209 wfg
~ % cat wfg
egrep -oi \\b[a-z]+|tr A-Z a-z|egrep -wv 'the|and|of|to|a|i|it|in|or|is'|sort|uniq -c|sort -nr|head -22|perl -lape'($f,$w)=@F;$.>1or($q,$x)=($f,76-length$w);$b="_"x($f/$q*$x);$_="|$b| $w ";$.>1or$_=" $b\n$_"'
~ % # usage:
~ % sh wfg < 11.txt

~~hm，刚见上文：sort -nr - ＆gt; sort -n然后head - ＆gt; tail =＆gt; 208：）~~
update2：嗯，当然上面是愚蠢的，因为它会被颠倒过来。那么，209 update3：优化了排除regexp - ＆gt; 206个

egrep -oi \\b[a-z]+|tr A-Z a-z|egrep -wv 'the|and|o[fr]|to|a|i[tns]?'|sort|uniq -c|sort -nr|head -22|perl -lape'($f,$w)=@F;$.>1or($q,$x)=($f,76-length$w);$b="_"x($f/$q*$x);$_="|$b| $w ";$.>1or$_=" $b\n$_"'

为了好玩，这里是一个perl-only版本（更快）：

~ % wc -c pgolf
204 pgolf
~ % cat pgolf
perl -lne'$1=~/^(the|and|o[fr]|to|.|i[tns])$/i||$f{lc$1}++while/\b([a-z]+)/gi}{@w=(sort{$f{$b}<=>$f{$a}}keys%f)[0..21];$Q=$f{$_=$w[0]};$B=76-y///c;print" "."_"x$B;print"|"."_"x($B*$f{$_}/$Q)."| $_"for@w'
~ % # usage:
~ % sh pgolf < 11.txt

Answer 6

Ruby 207 213 211 210 207 203 <罢工> 201 200个字符

对Anurag的改进，纳入了rfusca的建议。同时删除了排序和其他一些小型高尔夫球的参数。

w=(STDIN.read.downcase.scan(/[a-z]+/)-%w{the and of to a i it in or is}).group_by{|x|x}.map{|x,y|[-y.size,x]}.sort.take 22;k,l=w[0];m=76.0-l.size;puts' '+'_'*m;w.map{|f,x|puts"|#{'_'*(m*f/k)}| #{x} "}

执行为：

ruby GolfedWordFrequencies.rb < Alice.txt

编辑：将'puts'放回去，需要在那里避免输出中有引号编辑2：更改文件 - ＆gt; IO
Edit3：删除/ i
编辑4：删除括号（f * 1.0），重述
Edit5：为第一行使用字符串添加;在原地扩展s 编辑6：制作浮动，删除1.0。编辑：不起作用，改变长度。编辑：没有比以前更糟了编辑7：使用STDIN.read。

Answer 7

Mathematica（ 297 284 248 244 242 199 chars）Pure Functional

和Zipf的法律测试

看妈妈......没有变化，没有手，没有头

编辑1＆gt;定义了一些短号（284个字符）

f[x_, y_] := Flatten[Take[x, All, y]]; 

BarChart[f[{##}, -1], 
         BarOrigin -> Left, 
         ChartLabels -> Placed[f[{##}, 1], After], 
         Axes -> None
] 
& @@
Take[
  SortBy[
     Tally[
       Select[
        StringSplit[ToLowerCase[Import[i]], RegularExpression["\\W+"]], 
       !MemberQ[{"the", "and", "of", "to", "a", "i", "it", "in", "or","is"}, #]&]
     ], 
  Last], 
-22]

一些解释

Import[] 
   # Get The File

ToLowerCase []
   # To Lower Case :)

StringSplit[ STRING , RegularExpression["\\W+"]]
   # Split By Words, getting a LIST

Select[ LIST, !MemberQ[{LIST_TO_AVOID}, #]&]
   #  Select from LIST except those words in LIST_TO_AVOID
   #  Note that !MemberQ[{LIST_TO_AVOID}, #]& is a FUNCTION for the test

Tally[LIST]
   # Get the LIST {word,word,..} 
     and produce another  {{word,counter},{word,counter}...}

SortBy[ LIST ,Last]
   # Get the list produced bt tally and sort by counters
     Note that counters are the LAST element of {word,counter}

Take[ LIST ,-22]
   # Once sorted, get the biggest 22 counters

BarChart[f[{##}, -1], ChartLabels -> Placed[f[{##}, 1], After]] &@@ LIST
   # Get the list produced by Take as input and produce a bar chart

f[x_, y_] := Flatten[Take[x, All, y]]
   # Auxiliary to get the list of the first or second element of lists of lists x_
     dependending upon y
   # So f[{##}, -1] is the list of counters
   # and f[{##}, 1] is the list of words (labels for the chart)

输出

alt text http://i49.tinypic.com/2n8mrer.jpg

Mathematica不太适合打高尔夫球，这只是因为长而描述的功能名称。像“RegularExpression []”或“StringSplit []”这样的函数让我哭泣:(。

Zipf的法律测试

Zipf's law预测，对于自然语言文字，日志（排名）与日志（出现次数）的关键字会出现线性关系。

该法用于开发用于密码学和数据压缩的算法。（但它不是LZW算法中的“Z”）。

在我们的文本中，我们可以使用以下
进行测试
f[x_, y_] := Flatten[Take[x, All, y]]; ListLogLogPlot[ Reverse[f[{##}, -1]], AxesLabel -> {"Log (Rank)", "Log Counter"}, PlotLabel -> "Testing Zipf's Law"] & @@ Take[ SortBy[ Tally[ StringSplit[ToLowerCase[b], RegularExpression["\\W+"]] ], Last], -1000]

结果是（非常线性）

alt text http://i46.tinypic.com/33fcmdk.jpg

编辑6＆gt; （242 Chars）

重构正则表达式（不再选择功能）
删掉1个字符
功能“f”的更有效定义

f = Flatten[Take[#1, All, #2]]&; BarChart[ f[{##}, -1], BarOrigin -> Left, ChartLabels -> Placed[f[{##}, 1], After], Axes -> None] & @@ Take[ SortBy[ Tally[ StringSplit[ToLowerCase[Import[i]], RegularExpression["(\\W|\\b(.|the|and|of|to|i[tns]|or)\\b)+"]] ], Last], -22]

编辑7→199个字符

BarChart[#2, BarOrigin->Left, ChartLabels->Placed[#1, After], Axes->None]&@@ Transpose@Take[SortBy[Tally@StringSplit[ToLowerCase@Import@i, RegularExpression@"(\\W|\\b(.|the|and|of|to|i[tns]|or)\\b)+"],Last], -22]

用f和Transpose（Slot / #1）参数替换#2。

我们不需要stinkin'括号（尽可能使用f@x代替f[x]）

Answer 8

C＃ - 510 451 436 446 434 426 < / strike> 422 chars（minified）

不是那么短，但现在可能是正确的！注意，以前的版本没有显示条形的第一行，没有正确缩放条形，下载文件而不是从stdin获取它，并且没有包含所有必需的C＃详细程度。如果C＃不需要那么多额外的废话，你可以很容易地刮掉很多笔画。也许Powershell会做得更好。

using C=System.Console; // alias for Console using System.Linq; // for Split, GroupBy, Select, OrderBy, etc. class Class // must define a class { static void Main() // must define a Main { // split into words var allwords = System.Text.RegularExpressions.Regex.Split( // convert stdin to lowercase C.In.ReadToEnd().ToLower(), // eliminate stopwords and non-letters @"(?:\b(?:the|and|of|to|a|i[tns]?|or)\b|\W)+") .GroupBy(x => x) // group by words .OrderBy(x => -x.Count()) // sort descending by count .Take(22); // take first 22 words // compute length of longest bar + word var lendivisor = allwords.Max(y => y.Count() / (76.0 - y.Key.Length)); // prepare text to print var toPrint = allwords.Select(x=> new { // remember bar pseudographics (will be used in two places) Bar = new string('_',(int)(x.Count()/lendivisor)), Word=x.Key }) .ToList(); // convert to list so we can index into it // print top of first bar C.WriteLine(" " + toPrint[0].Bar); toPrint.ForEach(x => // for each word, print its bar and the word C.WriteLine("|" + x.Bar + "| " + x.Word)); } }

带有 lendivisor 的422个字符内联（这使得它慢22倍）在下面的表单中（用于选择空格的换行符）：

using System.Linq;using C=System.Console;class M{static void Main(){var a=System.Text.RegularExpressions.Regex.Split(C.In.ReadToEnd().ToLower(),@"(?:\b(?:the|and|of|to|a|i[tns]?|or)\b|\W)+").GroupBy(x=>x).OrderBy(x=>-x.Count()).Take(22);var b=a.Select(x=>new{p=new string('_',(int)(x.Count()/a.Max(y=>y.Count()/(76d-y.Key.Length)))),t=x.Key}).ToList();C.WriteLine(" "+b[0].p);b.ForEach(x=>C.WriteLine("|"+x.p+"| "+x.t));}}

Answer 9

Perl， 237 229 209 chars

（再次更新以使用更多肮脏的高尔夫技巧击败Ruby版本，将split/[^a-z/,lc替换为lc=~/[a-z]+/g，并取消检查另一个地方的空字符串。这些受到Ruby版本的启发，所以信用到期的信用。）

更新：现在使用Perl 5.10！将print替换为say，并使用~~来避免map。必须在命令行上调用perl -E '<one-liner>' alice.txt。由于整个脚本都在一行上，因此将其写成一行应该不会有任何困难：）。

 @s=qw/the and of to a i it in or is/;$c{$_}++foreach grep{!($_~~@s)}map{lc=~/[a-z]+/g}<>;@s=sort{$c{$b}<=>$c{$a}}keys%c;$f=76-length$s[0];say" "."_"x$f;say"|"."_"x($c{$_}/$c{$s[0]}*$f)."| $_ "foreach@s[0..21];

请注意，此版本针对大小写进行了规范化。这不会缩短任何解决方案，因为删除,lc（对于较低的大小）需要您将A-Z添加到拆分正则表达式中，因此它是一个清洗。

如果你的系统中换行符是一个字符而不是两个字符，则可以使用文字换行符代替\n将其缩短另外两个字符。但是，我没有这样写过上面的样本，因为它更“清楚”（哈！）那样。

这是一个大致正确但不够远的perl解决方案：

use strict;
use warnings;

my %short = map { $_ => 1 } qw/the and of to a i it in or is/;
my %count = ();

$count{$_}++ foreach grep { $_ && !$short{$_} } map { split /[^a-zA-Z]/ } (<>);
my @sorted = (sort { $count{$b} <=> $count{$a} } keys %count)[0..21];
my $widest = 76 - (length $sorted[0]);

print " " . ("_" x $widest) . "\n";
foreach (@sorted)
{
    my $width = int(($count{$_} / $count{$sorted[0]}) * $widest);
    print "|" . ("_" x $width) . "| $_ \n";
}

以下内容尽可能短，同时保持相对可读性。（392个字符）。

%short = map { $_ => 1 } qw/the and of to a i it in or is/;
%count;

$count{$_}++ foreach grep { $_ && !$short{$_} } map { split /[^a-z]/, lc } (<>);
@sorted = (sort { $count{$b} <=> $count{$a} } keys %count)[0..21];
$widest = 76 - (length $sorted[0]);

print " " . "_" x $widest . "\n";
print"|" . "_" x int(($count{$_} / $count{$sorted[0]}) * $widest) . "| $_ \n" foreach @sorted;

Answer 10

Windows PowerShell，199个字符

$x=$input-split'\P{L}'-notmatch'^(the|and|of|to|.?|i[tns]|or)$'|group|sort *
filter f($w){' '+'_'*$w
$x[-1..-22]|%{"|$('_'*($w*$_.Count/$x[-1].Count))| "+$_.Name}}
f(76..1|?{!((f $_)-match'.'*80)})[0]

（最后一个换行符不是必需的，但为了便于阅读，此处包括在内。）

（当前代码和我的测试文件可用in my SVN repository。我希望我的测试用例能够捕获最常见的错误（条形长度，正则表达式匹配问题以及其他一些错误）。

假设：

US ASCII作为输入。使用Unicode可能会很奇怪。
文本

两个

History

轻松版（137），因为现在已经单独计算了，显然：

($x=$input-split'\P{L}'-notmatch'^(the|and|of|to|.?|i[tns]|or)$'|group|sort *)[-1..-22]|%{"|$('_'*(76*$_.Count/$x[-1].Count))| "+$_.Name}

没有关闭第一个栏
不考虑非首字的字长

与其他解决方案相比，一个字符的条形长度的变化是由于PowerShell在将浮点数转换为整数时使用舍入而不是截断。由于任务只需要比例长度，这应该没问题。

与其他解决方案相比，我采用了稍微不同的方法来确定最长的条形长度，只需尝试并获取最高的长度，其中没有行超过80个字符。

可以找到解释的旧版本here。

Answer 11

Ruby，215， 216 ， 218 ， 221 ， 224 ， 236 ， 237 chars

更新1： Hurray ！它与JS Bangs'solution相关联。想不出再削减的方法了：）

更新2：玩了一个肮脏的高尔夫球技巧。已将each更改为map以保存1个字符：）

更新3：将File.read更改为IO.read +2。 Array.group_by效果不佳，改为reduce +6。在regex +1中使用downcase的下套管后不需要不区分大小写的检查。通过否定值+6可以轻松地按降序排序。总节省+15

更新4：[0]而不是.first，+ 3。（@Shtééf）

更新5：就地扩展变量l，+ 1。在原地扩展变量s，+ 2。（@Shtééf）

更新6：对第一行+2使用字符串加法而不是插值。（@Shtééf）

w=(IO.read($_).downcase.scan(/[a-z]+/)-%w{the and of to a i it in or is}).reduce(Hash.new 0){|m,o|m[o]+=1;m}.sort_by{|k,v|-v}.take 22;m=76-w[0][0].size;puts' '+'_'*m;w.map{|x,f|puts"|#{'_'*(f*1.0/w[0][1]*m)}| #{x} "}

更新7：我经历了大量的喧嚣，使用实例变量检测循环中的第一次迭代循环。我得到的只是+1，尽管可能有潜力。保留以前的版本，因为我相信这一个是黑魔法。（@Shtééf）

(IO.read($_).downcase.scan(/[a-z]+/)-%w{the and of to a i it in or is}).reduce(Hash.new 0){|m,o|m[o]+=1;m}.sort_by{|k,v|-v}.take(22).map{|x,f|@f||(@f=f;puts' '+'_'*(@m=76-x.size));puts"|#{'_'*(f*1.0/@f*@m)}| #{x} "}

可读版本

string = File.read($_).downcase

words = string.scan(/[a-z]+/i)
allowed_words = words - %w{the and of to a i it in or is}
sorted_words = allowed_words.group_by{ |x| x }.map{ |x,y| [x, y.size] }.sort{ |a,b| b[1] <=> a[1] }.take(22)
highest_frequency = sorted_words.first
highest_frequency_count = highest_frequency[1]
highest_frequency_word = highest_frequency[0]

word_length = highest_frequency_word.size
widest = 76 - word_length

puts " #{'_' * widest}"    
sorted_words.each do |word, freq|
  width = (freq * 1.0 / highest_frequency_count) * widest
  puts "|#{'_' * width}| #{word} "
end

使用：

echo "Alice.txt" | ruby -ln GolfedWordFrequencies.rb

输出：

 _________________________________________________________________________
|_________________________________________________________________________| she 
|_______________________________________________________________| you 
|____________________________________________________________| said 
|_____________________________________________________| alice 
|_______________________________________________| was 
|___________________________________________| that 
|____________________________________| as 
|________________________________| her 
|_____________________________| with 
|_____________________________| at 
|____________________________| s 
|____________________________| t 
|__________________________| on 
|__________________________| all 
|_______________________| this 
|_______________________| for 
|_______________________| had 
|_______________________| but 
|______________________| be 
|_____________________| not 
|____________________| they 
|____________________| so

Answer 12

Python 2.x，自由主义方法= 227 183 chars

import sys,re
t=re.split('\W+',sys.stdin.read().lower())
r=sorted((-t.count(w),w)for w in set(t)if w not in'andithetoforinis')[:22]
for l,w in r:print(78-len(r[0][1]))*l/r[0][0]*'=',w

在实现中允许自由，我构建了一个字符串连接，其中包含要求排除的所有单词（the, and, of, to, a, i, it, in, or, is） - 此外它还排除了两个臭名昭着的“单词”s和{{1从示例中 - 我免费提供t的排除。我尝试将这些词语的所有连接与爱丽丝，詹姆斯国王的圣经和术语文件中的单词语料库联系起来，以查看是否有任何单词会被字符串误排除。这就是我以两个排除字符串结束的方式：an, for, he和itheandtoforinis。

PS。借用其他解决方案来缩短代码。

andithetoforinis

咆哮

关于要忽略的词语，人们会认为这些词汇将取自英语中最常用词汇的列表。该列表取决于使用的text corpus。根据一个最受欢迎的列表（http://en.wikipedia.org/wiki/Most_common_words_in_English，http://www.english-for-students.com/Frequently-Used-Words.html，http://www.sporcle.com/games/common_english_words.php），前10个字是：=========================================================================== she ================================================================= you ============================================================== said ====================================================== alice ================================================ was ============================================ that ===================================== as ================================= her ============================== at ============================== with =========================== on =========================== all ======================== this ======================== had ======================= but ====================== be ====================== not ===================== they ==================== so =================== very =================== what ================= little

爱丽丝梦游仙境文字中的前10个单词是the be(am/are/is/was/were) to of and a in that have I
术语文本（v4.4.7）中的前10个单词是the and to a of it she i you said

所以，问题是为什么the a of to and in is that or for被包含在问题的忽略列表中，当or（使用率最高的第8个）不是时，它的受欢迎程度约为30。因此我认为应该动态地提供忽略列表（或者可以省略）。

替代想法只是跳过结果中的前10个单词 - 这实际上会缩短解决方案（初级 - 只需要显示第11到第32个条目）。

Python 2.x，punctilious approach = 277 243个字符

上面代码中绘制的图表是简化的（条形图只使用一个字符）。如果想要从问题描述中完全重现图表（这不是必需的），则此代码将执行此操作：

that

我对10个单词的随机选择有问题，要排除import sys,re t=re.split('\W+',sys.stdin.read().lower()) r=sorted((-t.count(w),w)for w in set(t)-set(sys.argv))[:22] h=min(9*l/(77-len(w))for l,w in r) print'',9*r[0][0]/h*'_' for l,w in r:print'|'+9*l/h*'_'+'|',w，所以这些将作为命令行参数传递，如下所示：
the, and, of, to, a, i, it, in, or, is

如果我们考虑在命令行= 243上传递的“原始”忽略列表，这是213个字符+30

PS。第二个代码也对所有顶部单词的长度进行“调整”，因此在退化情况下它们都不会溢出。

python WordFrequencyChart.py  the and of to a i it in or is  <"Alice's Adventures in Wonderland.txt"

Answer 13

Haskell - 366 351 344 337 333个字符

（main中的一个换行符为可读性添加，并且在最后一行末尾不需要换行符。）

import Data.List
import Data.Char
l=length
t=filter
m=map
f c|isAlpha c=toLower c|0<1=' '
h w=(-l w,head w)
x!(q,w)='|':replicate(minimum$m(q?)x)'_'++"| "++w
q?(g,w)=q*(77-l w)`div`g
b x=m(x!)x
a(l:r)=(' ':t(=='_')l):l:r
main=interact$unlines.a.b.take 22.sort.m h.group.sort
  .t(`notElem`words"the and of to a i it in or is").words.m f

通过向后阅读interact的参数，可以最好地看到它的工作原理：

map f小写字母，用空格替换其他所有内容。
words生成一个单词列表，删除分隔的空格。
filter ( notElem words "the and of to a i it in or is")会丢弃所有带有禁词的条目。
group . sort对单词进行排序，并将相同的单词分组到列表中。
map h将每个相同字词列表映射到(-frequency, word)形式的元组。
take 22 . sort按降序频率（第一个元组条目）对元组进行排序，并仅保留前22个元组。
b将元组映射到条形图（见下文）。
a在第一行下划线之前，以完成最顶部的栏。
unlines将所有这些行与新行一起加入。

棘手的一点就是让杆长正确。我假设只有下划线计算到条的长度，所以||将是一个零长度的条。函数b将c x映射到x，其中x是直方图列表。整个列表将传递给c，因此c的每次调用都可以通过调用u来计算自身的比例因子。通过这种方式，我避免使用浮点数学或有理数，其转换函数和导入会占用很多字符。

请注意使用-frequency的技巧。这样就不需要reverse sort，因为排序（升序）-frequency将首先放置频率最高的字词。之后，在函数u中，两个-frequency值相乘，这将取消否定。

Answer 14

Python 3.1 - 245 229个字符

我想使用Counter是一种欺骗行为:)我刚刚在一周前阅读过它，所以这是查看它是如何工作的绝佳机会。

import re,collections
o=collections.Counter([w for w in re.findall("[a-z]+",open("!").read().lower())if w not in"a and i in is it of or the to".split()]).most_common(22)
print('\n'.join('|'+76*v//o[0][1]*'_'+'| '+k for k,v in o))

打印出来：

|____________________________________________________________________________| she
|__________________________________________________________________| you
|_______________________________________________________________| said
|_______________________________________________________| alice
|_________________________________________________| was
|_____________________________________________| that
|_____________________________________| as
|__________________________________| her
|_______________________________| with
|_______________________________| at
|______________________________| s
|_____________________________| t
|____________________________| on
|___________________________| all
|________________________| this
|________________________| for
|________________________| had
|________________________| but
|______________________| be
|______________________| not
|_____________________| they
|____________________| so

部分代码是从AKX的解决方案中“借用”的。

Answer 15

PHP CLI版本（450个字符）

这个解决方案考虑了大多数纯粹主义者方便地选择忽略的最后一个要求。花费了170个字符！

用法：php.exe <this.php> <file.txt>

缩小的：

<?php $a=array_count_values(array_filter(preg_split('/[^a-z]/',strtolower(file_get_contents($argv[1])),-1,1),function($x){return !preg_match("/^(.|the|and|of|to|it|in|or|is)$/",$x);}));arsort($a);$a=array_slice($a,0,22);function R($a,$F,$B){$r=array();foreach($a as$x=>$f){$l=strlen($x);$r[$x]=$b=$f*$B/$F;if($l+$b>76)return R($a,$f,76-$l);}return$r;}$c=R($a,max($a),76-strlen(key($a)));foreach($a as$x=>$f)echo '|',str_repeat('-',$c[$x]),"| $x\n";?>

人类可读：

<?php

// Read:
$s = strtolower(file_get_contents($argv[1]));

// Split:
$a = preg_split('/[^a-z]/', $s, -1, PREG_SPLIT_NO_EMPTY);

// Remove unwanted words:
$a = array_filter($a, function($x){
       return !preg_match("/^(.|the|and|of|to|it|in|or|is)$/",$x);
     });

// Count:
$a = array_count_values($a);

// Sort:
arsort($a);

// Pick top 22:
$a=array_slice($a,0,22);


// Recursive function to adjust bar widths
// according to the last requirement:
function R($a,$F,$B){
    $r = array();
    foreach($a as $x=>$f){
        $l = strlen($x);
        $r[$x] = $b = $f * $B / $F;
        if ( $l + $b > 76 )
            return R($a,$f,76-$l);
    }
    return $r;
}

// Apply the function:
$c = R($a,max($a),76-strlen(key($a)));


// Output:
foreach ($a as $x => $f)
    echo '|',str_repeat('-',$c[$x]),"| $x\n";

?>

输出：

|-------------------------------------------------------------------------| she
|---------------------------------------------------------------| you
|------------------------------------------------------------| said
|-----------------------------------------------------| alice
|-----------------------------------------------| was
|-------------------------------------------| that
|------------------------------------| as
|--------------------------------| her
|-----------------------------| at
|-----------------------------| with
|--------------------------| on
|--------------------------| all
|-----------------------| this
|-----------------------| for
|-----------------------| had
|-----------------------| but
|----------------------| be
|---------------------| not
|--------------------| they
|--------------------| so
|-------------------| very
|------------------| what

当有一个长字时，条形图会正确调整：

|--------------------------------------------------------| she
|---------------------------------------------------| thisisareallylongwordhere
|-------------------------------------------------| you
|-----------------------------------------------| said
|-----------------------------------------| alice
|------------------------------------| was
|---------------------------------| that
|---------------------------| as
|-------------------------| her
|-----------------------| with
|-----------------------| at
|--------------------| on
|--------------------| all
|------------------| this
|------------------| for
|------------------| had
|-----------------| but
|-----------------| be
|----------------| not
|---------------| they
|---------------| so
|--------------| very

Answer 16

perl， 205 191 189个字符/ 205个字符（完全实现）

有些部分受到早期perl / ruby提交的启发，一些类似的想法是独立到达的，其他部分是原创的。较短的版本还包含了我从其他提交中看到/学到的一些内容。

原件：

$k{$_}++for grep{$_!~/^(the|and|of|to|a|i|it|in|or|is)$/}map{lc=~/[a-z]+/g}<>;@t=sort{$k{$b}<=>$k{$a}}keys%k;$l=76-length$t[0];printf" %s
",'_'x$l;printf"|%s| $_
",'_'x int$k{$_}/$k{$t[0]}*$l for@t[0..21];

~~最新版本~~ 191个字符：

/^(the|and|of|to|.|i[tns]|or)$/||$k{$_}++for map{lc=~/[a-z]+/g}<>;@e=sort{$k{$b}<=>$k{$a}}keys%k;$n=" %s
";$r=(76-y///c)/$k{$_=$e[0]};map{printf$n,'_'x($k{$_}*$r),$_;$n="|%s| %s
"}@e[0,0..21]

最新版本低至189个字符：

/^(the|and|of|to|.|i[tns]|or)$/||$k{$_}++for map{lc=~/[a-z]+/g}<>;@_=sort{$k{$b}<=>$k{$a}}keys%k;$n=" %s
";$r=(76-m//)/$k{$_=$_[0]};map{printf$n,'_'x($k{$_}*$r),$_;$n="|%s| %s
"}@_[0,0..21]

此版本（205个字符）说明的字词长于以后找到的字。

/^(the|and|of|to|.|i[tns]|or)$/||$k{$_}++for map{lc=~/[a-z]+/g}<>;($r)=sort{$a<=>$b}map{(76-y///c)/$k{$_}}@e=sort{$k{$b}<=>$k{$a}}keys%k;$n=" %s
";map{printf$n,'_'x($k{$_}*$r),$_;$n="|%s| %s
";}@e[0,0..21]

Answer 17

JavaScript 1.8（SpiderMonkey） - 354

x={};p='|';e=' ';z=[];c=77
while(l=readline())l.toLowerCase().replace(/\b(?!(the|and|of|to|a|i[tns]?|or)\b)\w+/g,function(y)x[y]?x[y].c++:z.push(x[y]={w:y,c:1}))
z=z.sort(function(a,b)b.c-a.c).slice(0,22)
for each(v in z){v.r=v.c/z[0].c
c=c>(l=(77-v.w.length)/v.r)?l:c}for(k in z){v=z[k]
s=Array(v.r*c|0).join('_')
if(!+k)print(e+s+e)
print(p+s+p+e+v.w)}

可悲的是，Rhino版本中的for([k,v]in z)似乎不想在SpiderMonkey中工作，readFile()比使用readline()更容易，但升级到1.8允许我们使用函数闭包来减少几行...

添加空格以提高可读性：

x={};p='|';e=' ';z=[];c=77
while(l=readline())
  l.toLowerCase().replace(/\b(?!(the|and|of|to|a|i[tns]?|or)\b)\w+/g,
   function(y) x[y] ? x[y].c++ : z.push( x[y] = {w: y, c: 1} )
  )
z=z.sort(function(a,b) b.c - a.c).slice(0,22)
for each(v in z){
  v.r=v.c/z[0].c
  c=c>(l=(77-v.w.length)/v.r)?l:c
}
for(k in z){
  v=z[k]
  s=Array(v.r*c|0).join('_')
  if(!+k)print(e+s+e)
  print(p+s+p+e+v.w)
}

用法： js golf.js < input.txt

<强>输出：

 _________________________________________________________________________ 
|_________________________________________________________________________| she
|_______________________________________________________________| you
|____________________________________________________________| said
|____________________________________________________| alice
|______________________________________________| was
|___________________________________________| that
|___________________________________| as
|________________________________| her
|_____________________________| at
|_____________________________| with
|____________________________| s
|____________________________| t
|__________________________| on
|_________________________| all
|_______________________| this
|______________________| for
|______________________| had
|______________________| but
|_____________________| be
|_____________________| not
|___________________| they
|___________________| so

（基本版 - 不能正确处理条宽）

JavaScript（Rhino） - 405 395 387 377 368 343 304个字符

~~我认为我的排序逻辑已关闭，但是......我duno。~~ Brainfart已修复。

缩小（有时滥用\n被解释为;）：

x={};p='|';e=' ';z=[]
readFile(arguments[0]).toLowerCase().replace(/\b(?!(the|and|of|to|a|i[tns]?|or)\b)\w+/g,function(y){x[y]?x[y].c++:z.push(x[y]={w:y,c:1})})
z=z.sort(function(a,b){return b.c-a.c}).slice(0,22)
for([k,v]in z){s=Array((v.c/z[0].c)*70|0).join('_')
if(!+k)print(e+s+e)
print(p+s+p+e+v.w)}

Answer 18

Perl： 203 202 201 198 195 208 203/231 chars

$/=\0;/^(the|and|of|to|.|i[tns]|or)$/i||$x{lc$_}++for<>=~/[a-z]+/gi;map{$z=$x{$_};$y||{$y=(76-y///c)/$z}&&warn" "."_"x($z*$y)."\n";printf"|%.78s\n","_"x($z*$y)."| $_"}(sort{$x{$b}<=>$x{$a}}keys%x)[0..21]

备选，全面实施，包括指示行为（全局条形压缩）的病理案例，其中次要词既受欢迎又长到足以结合超过80个字符（此实施是231个字符 ）：

$/=\0;/^(the|and|of|to|.|i[tns]|or)$/i||$x{lc$_}++for<>=~/[a-z]+/gi;@e=(sort{$x{$b}<=>$x{$a}}keys%x)[0..21];for(@e){$p=(76-y///c)/$x{$_};($y&&$p>$y)||($y=$p)}warn" "."_"x($x{$e[0]}*$y)."\n";for(@e){warn"|"."_"x($x{$_}*$y)."| $_\n"}

规范没有说明必须转到STDOUT，因此我使用了perl的warn（）而不是print - 在那里保存了四个字符。使用map而不是foreach，但我觉得在split（join（））中仍然可以节省更多。仍然，把它降到203 - 可能会睡在它上面。至少Perl现在处于“shell，grep，tr，grep，sort，uniq，sort，head，perl”字样下，现在可用;）

PS：Reddit说“嗨”;）

更新：删除了join（），支持赋值和隐式标量转换连接。低至202.另请注意我已经利用可选的“忽略1个字母的单词”规则来关闭2个字符，因此请记住频率计数将反映这一点。

更新2：使用＆lt;＆gt;删除分配和隐式联接以杀死$ /以一次吞咽获取文件首先。相同的尺寸，但更肮脏。如果（$ $ y）{}为$ y || {}＆amp;＆amp ;,则省略另外1个char =＆gt; 201。

更新3：通过将lc移出地图块来早期控制小写（lc＆lt;＆gt;） - 将两个正则表达式交换为不再使用/ i选项，因为不再需要。传统的perlgolf ||交换显式条件x？y：z构造隐式条件构造 - /^...$/i?1:$x{$}++ for /^...$/||$x{$ }++已保存三个字符！ =＆GT; 198，打破了200障碍。可能很快就会睡觉......也许吧。

更新4：睡眠剥夺让我精神错乱。好。更疯狂。确定这只需要解析正常的快乐文本文件，如果它达到空值，我会放弃它。保存了两个字符。将“长度”替换为1-char更短（更多高尔夫球）y /// c - 你听到我，GolfScript ??我来找你!!! SOB

更新5：Sleep dep让我忘记了22row限制和后续行限制。回复到208处理。还不错，处理它的13个字符不是世界末日。使用perl的正则表达式内联eval，但无法将它同时工作和保存字符...大声笑。更新了示例以匹配当前输出。

更新6：删除了不需要的大括号保护（...），因为语法糖果++允许快乐地推动它。感谢Chas的意见。欧文斯（提醒我疲惫的大脑），在那里得到了人物类别[tns]解决方案。回到203。

更新7：添加了第二项工作，完全实现规范（包括次要长词的完整条形压缩行为，而不是大多数人正在做的截断，基于原始规范而没有病态示例情况）

<强>示例：

 _________________________________________________________________________
|_________________________________________________________________________| she
|_______________________________________________________________| you
|____________________________________________________________| said
|_____________________________________________________| alice
|_______________________________________________| was
|___________________________________________| that
|____________________________________| as
|________________________________| her
|_____________________________| with
|_____________________________| at
|__________________________| on
|__________________________| all
|_______________________| this
|_______________________| for
|_______________________| had
|_______________________| but
|______________________| be
|_____________________| not
|____________________| they
|____________________| so
|___________________| very
|__________________| what

病态案例中的替代实施：

 _______________________________________________________________
|_______________________________________________________________| she
|_______________________________________________________| superlongstringstring
|____________________________________________________| said
|______________________________________________| alice
|________________________________________| was
|_____________________________________| that
|_______________________________| as
|____________________________| her
|_________________________| with
|_________________________| at
|_______________________| on
|______________________| all
|____________________| this
|____________________| for
|____________________| had
|____________________| but
|___________________| be
|__________________| not
|_________________| they
|_________________| so
|________________| very
|________________| what

Answer 19

F＃，452个字符

明确：获取字数对a的序列，找到每列最佳字数乘数k，然后打印结果。

let a=
 stdin.ReadToEnd().Split(" .?!,\":;'\r\n".ToCharArray(),enum 1)
 |>Seq.map(fun s->s.ToLower())|>Seq.countBy id
 |>Seq.filter(fun(w,n)->not(set["the";"and";"of";"to";"a";"i";"it";"in";"or";"is"].Contains w))
 |>Seq.sortBy(fun(w,n)-> -n)|>Seq.take 22
let k=a|>Seq.map(fun(w,n)->float(78-w.Length)/float n)|>Seq.min
let u n=String.replicate(int(float(n)*k)-2)"_"
printfn" %s "(u(snd(Seq.nth 0 a)))
for(w,n)in a do printfn"|%s| %s "(u n)w

示例（我有不同的频率计数，不确定原因）：

% app.exe < Alice.txt

 _________________________________________________________________________
|_________________________________________________________________________| she
|_______________________________________________________________| you
|_____________________________________________________________| said
|_____________________________________________________| alice
|_______________________________________________| was
|___________________________________________| that
|___________________________________| as
|________________________________| her
|_____________________________| with
|_____________________________| at
|____________________________| t
|____________________________| s
|__________________________| on
|_________________________| all
|_______________________| this
|______________________| had
|______________________| for
|_____________________| but
|_____________________| be
|____________________| not
|___________________| they
|__________________| so

Answer 20

Python 2.6,347字符

import re
W,x={},"a and i in is it of or the to".split()
[W.__setitem__(w,W.get(w,0)-1)for w in re.findall("[a-z]+",file("11.txt").read().lower())if w not in x]
W=sorted(W.items(),key=lambda p:p[1])[:22]
bm=(76.-len(W[0][0]))/W[0][1]
U=lambda n:"_"*int(n*bm)
print "".join(("%s\n|%s| %s "%((""if i else" "+U(n)),U(n),w))for i,(w,n)in enumerate(W))

输出：

 _________________________________________________________________________
|_________________________________________________________________________| she 
|_______________________________________________________________| you 
|____________________________________________________________| said 
|_____________________________________________________| alice 
|_______________________________________________| was 
|___________________________________________| that 
|____________________________________| as 
|________________________________| her 
|_____________________________| with 
|_____________________________| at 
|____________________________| s 
|____________________________| t 
|__________________________| on 
|__________________________| all 
|_______________________| this 
|_______________________| for 
|_______________________| had 
|_______________________| but 
|______________________| be 
|_____________________| not 
|____________________| they 
|____________________| so

Answer 21

通用LISP，670个字符

我是LISP的新手，这是尝试使用哈希表进行计数（所以可能不是最紧凑的方法）。

(flet((r()(let((x(read-char t nil)))(and x(char-downcase x)))))(do((c(
make-hash-table :test 'equal))(w NIL)(x(r)(r))y)((not x)(maphash(lambda
(k v)(if(not(find k '("""the""and""of""to""a""i""it""in""or""is"):test
'equal))(push(cons k v)y)))c)(setf y(sort y #'> :key #'cdr))(setf y
(subseq y 0(min(length y)22)))(let((f(apply #'min(mapcar(lambda(x)(/(-
76.0(length(car x)))(cdr x)))y))))(flet((o(n)(dotimes(i(floor(* n f)))
(write-char #\_))))(write-char #\Space)(o(cdar y))(write-char #\Newline)
(dolist(x y)(write-char #\|)(o(cdr x))(format t "| ~a~%"(car x))))))
(cond((char<= #\a x #\z)(push x w))(t(incf(gethash(concatenate 'string(
reverse w))c 0))(setf w nil)))))

可以运行，例如 cat alice.txt | clisp -C golf.lisp。

以可读形式

(flet ((r () (let ((x (read-char t nil)))
               (and x (char-downcase x)))))
  (do ((c (make-hash-table :test 'equal))  ; the word count map
       w y                                 ; current word and final word list
       (x (r) (r)))  ; iteration over all chars
       ((not x)

        ; make a list with (word . count) pairs removing stopwords
        (maphash (lambda (k v)
                   (if (not (find k '("" "the" "and" "of" "to"
                                      "a" "i" "it" "in" "or" "is")
                                  :test 'equal))
                       (push (cons k v) y)))
                 c)

        ; sort and truncate the list
        (setf y (sort y #'> :key #'cdr))
        (setf y (subseq y 0 (min (length y) 22)))

        ; find the scaling factor
        (let ((f (apply #'min
                        (mapcar (lambda (x) (/ (- 76.0 (length (car x)))
                                               (cdr x)))
                                y))))
          ; output
          (flet ((outx (n) (dotimes (i (floor (* n f))) (write-char #\_))))
             (write-char #\Space)
             (outx (cdar y))
             (write-char #\Newline)
             (dolist (x y)
               (write-char #\|)
               (outx (cdr x))
               (format t "| ~a~%" (car x))))))

       ; add alphabetic to current word, and bump word counter
       ; on non-alphabetic
       (cond
        ((char<= #\a x #\z)
         (push x w))
        (t
         (incf (gethash (concatenate 'string (reverse w)) c 0))
         (setf w nil)))))

Answer 22

Gawk - 336（最初为507）个字符

（在修复输出格式之后;修复收缩的东西;调整;再次调整;删除一个完全不必要的排序步骤;再次调整;再次（哎呀这个打破格式化）;再调整一些;取Matt的挑战我拼命地调整了更多;发现另外一个地方可以节省一些，但是给了两个回来修复条形长度的bug）

嘿嘿！我暂时领先于[Matt的JavaScript] [1]解决方案^{反击挑战！ ;）} 和 [AKX的python] [2]。

这个问题似乎引出了一种实现原生关联数组的语言，所以当然我选择了一个在它们上面有一组可怕的运算符。特别是，您无法控制awk提供哈希映射元素的顺序，因此我反复扫描整个映射以查找当前最多的项目，打印它并从阵列中删除它

这一切都非常低效，我所做的所有高尔夫球场也变得非常糟糕。

缩小的：

{gsub("[^a-zA-Z]"," ");for(;NF;NF--)a[tolower($NF)]++}
END{split("the and of to a i it in or is",b," ");
for(w in b)delete a[b[w]];d=1;for(w in a){e=a[w]/(78-length(w));if(e>d)d=e}
for(i=22;i;--i){e=0;for(w in a)if(a[w]>e)e=a[x=w];l=a[x]/d-2;
t=sprintf(sprintf("%%%dc",l)," ");gsub(" ","_",t);if(i==22)print" "t;
print"|"t"| "x;delete a[x]}}

换行符仅为清晰起见：它们不是必需的，不应计算在内。

输出：

$ gawk -f wordfreq.awk.min < 11.txt 
 _________________________________________________________________________
|_________________________________________________________________________| she
|_______________________________________________________________| you
|____________________________________________________________| said
|____________________________________________________| alice
|______________________________________________| was
|__________________________________________| that
|___________________________________| as
|_______________________________| her
|____________________________| with
|____________________________| at
|___________________________| s
|___________________________| t
|_________________________| on
|_________________________| all
|______________________| this
|______________________| for
|______________________| had
|_____________________| but
|____________________| be
|____________________| not
|___________________| they
|__________________| so
$ sed 's/you/superlongstring/gI' 11.txt | gawk -f wordfreq.awk.min
 ______________________________________________________________________
|______________________________________________________________________| she
|_____________________________________________________________| superlongstring
|__________________________________________________________| said
|__________________________________________________| alice
|____________________________________________| was
|_________________________________________| that
|_________________________________| as
|______________________________| her
|___________________________| with
|___________________________| at
|__________________________| s
|__________________________| t
|________________________| on
|________________________| all
|_____________________| this
|_____________________| for
|_____________________| had
|____________________| but
|___________________| be
|___________________| not
|__________________| they
|_________________| so

可读; 633个字符（原为949）：

{
    gsub("[^a-zA-Z]"," ");
    for(;NF;NF--)
    a[tolower($NF)]++
}
END{
    # remove "short" words
    split("the and of to a i it in or is",b," ");
    for (w in b) 
    delete a[b[w]];
    # Find the bar ratio
    d=1;
    for (w in a) {
    e=a[w]/(78-length(w));
    if (e>d)
        d=e
    }
    # Print the entries highest count first
    for (i=22; i; --i){               
    # find the highest count
    e=0;
    for (w in a) 
        if (a[w]>e)
        e=a[x=w];
        # Print the bar
    l=a[x]/d-2;
    # make a string of "_" the right length
    t=sprintf(sprintf("%%%dc",l)," ");
    gsub(" ","_",t);
    if (i==22) print" "t;
    print"|"t"| "x;
    delete a[x]
    }
}

Answer 23

* sh（+ curl），部分解决方案

这是不完整的，但对于它的地狱，这里的字频率计算问题的一半是192字节：

curl -s http://www.gutenberg.org/files/11/11.txt|sed -e 's@[^a-z]@\n@gi'|tr '[:upper:]' '[:lower:]'|egrep -v '(^[^a-z]*$|\b(the|and|of|to|a|i|it|in|or|is)\b)' |sort|uniq -c|sort -n|tail -n 22

Answer 24

Perl，185 char

~~200（稍微破碎）~~ <击> 199 <击> 197 <击> 195 <击> 193 <击> 187 185个字符。最后两个新行很重要。符合规范。

map$X{+lc}+=!/^(.|the|and|to|i[nst]|o[rf])$/i,/[a-z]+/gfor<>;
$n=$n>($:=$X{$_}/(76-y+++c))?$n:$:for@w=(sort{$X{$b}-$X{$a}}%X)[0..21];
die map{$U='_'x($X{$_}/$n);" $U
"x!$z++,"|$U| $_
"}@w

第一行将有效单词的计数加载到%X。

第二行计算最小比例因子，以便所有输出行都是<= 80个字符。

第三行（包含两个换行符）产生输出。

Answer 25

C（828）

它看起来很像混淆代码，并使用glib作为字符串，列表和哈希。 wc -m的字数统计 828 。它不考虑单字词。要计算条形的最大长度，它会考虑所有条中最长的单词，而不仅仅是前22条。这是否与规格有偏差？

它不处理故障，也不释放已用的内存。

#include <glib.h>
#define S(X)g_string_##X
#define H(X)g_hash_table_##X
GHashTable*h;int m,w=0,z=0;y(const void*a,const void*b){int*A,*B;A=H(lookup)(h,a);B=H(lookup)(h,b);return*B-*A;}void p(void*d,void*u){int *v=H(lookup)(h,d);if(w<22){g_printf("|");*v=*v*(77-z)/m;while(--*v>=0)g_printf("=");g_printf("| %s\n",d);w++;}}main(c){int*v;GList*l;GString*s=S(new)(NULL);h=H(new)(g_str_hash,g_str_equal);char*n[]={"the","and","of","to","it","in","or","is"};while((c=getchar())!=-1){if(isalpha(c))S(append_c)(s,tolower(c));else{if(s->len>1){for(c=0;c<8;c++)if(!strcmp(s->str,n[c]))goto x;if((v=H(lookup)(h,s->str))!=NULL)++*v;else{z=MAX(z,s->len);v=g_malloc(sizeof(int));*v=1;H(insert)(h,g_strdup(s->str),v);}}x:S(truncate)(s,0);}}l=g_list_sort(H(get_keys)(h),y);m=*(int*)H(lookup)(h,g_list_first(l)->data);g_list_foreach(l,p,NULL);}

Answer 26

Java - 886 865 756 744 742 744 < / strike> 752 742 714 680个字符

在第742次之前更新：改进了正则表达式，删除了多余的参数化类型，删除了多余的空格。

更新742＆gt; 744个字符：修复了固定长度的黑客攻击。它只依赖于第一个单词，而不是其他单词（尚未）。找到了几个缩短代码的地方（正则表达式\\s替换为而ArrayList替换为Vector）。我现在正在寻找一种简短的方法来删除Commons IO依赖并从stdin读取。

更新744＆gt; 752个字符：我删除了公共依赖项。它现在从stdin读取。将文本粘贴到标准输入并点击Ctrl+Z以获得结果。

更新752＆gt; 742个字符：我删除了public和一个空格，使得classname 1 char而不是2，现在它忽略了一个字母的单词。

更新742＆gt; 714个字符：根据Carl的评论更新：删除了冗余分配（742＆gt; 730），将m.containsKey(k)替换为m.get(k)!=null（730＆gt; 728），引入了线路的子串（728＆gt; ; 714）。

更新714＆gt; 680个字符：根据Rotsor的评论进行了更新：改进了条形码大小计算以删除不必要的转换，并改进了split()以删除不必要的replaceAll()。

import java.util.*;class F{public static void main(String[]a)throws Exception{StringBuffer b=new StringBuffer();for(int c;(c=System.in.read())>0;b.append((char)c));final Map<String,Integer>m=new HashMap();for(String w:b.toString().toLowerCase().split("(\\b(.|the|and|of|to|i[tns]|or)\\b|\\W)+"))m.put(w,m.get(w)!=null?m.get(w)+1:1);List<String>l=new Vector(m.keySet());Collections.sort(l,new Comparator(){public int compare(Object l,Object r){return m.get(r)-m.get(l);}});int c=76-l.get(0).length();String s=new String(new char[c]).replace('\0','_');System.out.println(" "+s);for(String w:l.subList(0,22))System.out.println("|"+s.substring(0,m.get(w)*c/m.get(l.get(0)))+"| "+w);}}

更易阅读的版本：

import java.util.*; class F{ public static void main(String[]a)throws Exception{ StringBuffer b=new StringBuffer();for(int c;(c=System.in.read())>0;b.append((char)c)); final Map<String,Integer>m=new HashMap();for(String w:b.toString().toLowerCase().split("(\\b(.|the|and|of|to|i[tns]|or)\\b|\\W)+"))m.put(w,m.get(w)!=null?m.get(w)+1:1); List<String>l=new Vector(m.keySet());Collections.sort(l,new Comparator(){public int compare(Object l,Object r){return m.get(r)-m.get(l);}}); int c=76-l.get(0).length();String s=new String(new char[c]).replace('\0','_');System.out.println(" "+s); for(String w:l.subList(0,22))System.out.println("|"+s.substring(0,m.get(w)*c/m.get(l.get(0)))+"| "+w); } }

输出：

_________________________________________________________________________ |_________________________________________________________________________| she |_______________________________________________________________| you |____________________________________________________________| said |_____________________________________________________| alice |_______________________________________________| was |___________________________________________| that |____________________________________| as |________________________________| her |_____________________________| with |_____________________________| at |__________________________| on |__________________________| all |_______________________| this |_______________________| for |_______________________| had |_______________________| but |______________________| be |_____________________| not |____________________| they |____________________| so |___________________| very |__________________| what

Java还没有String#join()和closures（还有）。

由Rotsor编辑：

我对您的解决方案进行了一些更改：

用String []
替换List
重用'args'参数而不是声明我自己的String数组。也用它作为.ToArray（）
的参数
用字符串替换StringBuffer（是的，是的，可怕的性能）

用早期暂停的选择排序替换Java排序（只需要找到前22个元素）

将一些int声明聚合到单个语句中

实现了非作弊算法，找到最有限的输出线。没有FP实现它。

修正了文本
中少于22个不同单词时程序崩溃的问题
实现了一种新的读取输入算法，该算法速度快，只比慢速输入长9个字符。

简明代码为 ~~688~~ ~~711~~ 684 个字符：

import java.util.*;class F{public static void main(String[]l)throws Exception{Map<String,Integer>m=new HashMap();String w="";int i=0,k=0,j=8,x,y,g=22;for(;(j=System.in.read())>0;w+=(char)j);for(String W:w.toLowerCase().split("(\\b(.|the|and|of|to|i[tns]|or)\\b|\\W)+"))m.put(W,m.get(W)!=null?m.get(W)+1:1);l=m.keySet().toArray(l);x=l.length;if(x<g)g=x;for(;i<g;++i)for(j=i;++j<x;)if(m.get(l[i])<m.get(l[j])){w=l[i];l[i]=l[j];l[j]=w;}for(;k<g;k++){x=76-l[k].length();y=m.get(l[k]);if(k<1||y*i>x*j){i=x;j=y;}}String s=new String(new char[m.get(l[0])*i/j]).replace('\0','_');System.out.println(" "+s);for(k=0;k<g;k++){w=l[k];System.out.println("|"+s.substring(0,m.get(w)*i/j)+"| "+w);}}}

快速版（ ~~720~~ 693 个字符）

import java.util.*;class F{public static void main(String[]l)throws Exception{Map<String,Integer>m=new HashMap();String w="";int i=0,k=0,j=8,x,y,g=22;for(;j>0;){j=System.in.read();if(j>90)j-=32;if(j>64&j<91)w+=(char)j;else{if(!w.matches("^(|.|THE|AND|OF|TO|I[TNS]|OR)$"))m.put(w,m.get(w)!=null?m.get(w)+1:1);w="";}}l=m.keySet().toArray(l);x=l.length;if(x<g)g=x;for(;i<g;++i)for(j=i;++j<x;)if(m.get(l[i])<m.get(l[j])){w=l[i];l[i]=l[j];l[j]=w;}for(;k<g;k++){x=76-l[k].length();y=m.get(l[k]);if(k<1||y*i>x*j){i=x;j=y;}}String s=new String(new char[m.get(l[0])*i/j]).replace('\0','_');System.out.println(" "+s);for(k=0;k<g;k++){w=l[k];System.out.println("|"+s.substring(0,m.get(w)*i/j)+"| "+w);}}}

更易阅读的版本：

import java.util.*;class F{public static void main(String[]l)throws Exception{ Map<String,Integer>m=new HashMap();String w=""; int i=0,k=0,j=8,x,y,g=22; for(;j>0;){j=System.in.read();if(j>90)j-=32;if(j>64&j<91)w+=(char)j;else{ if(!w.matches("^(|.|THE|AND|OF|TO|I[TNS]|OR)$"))m.put(w,m.get(w)!=null?m.get(w)+1:1);w=""; }} l=m.keySet().toArray(l);x=l.length;if(x<g)g=x; for(;i<g;++i)for(j=i;++j<x;)if(m.get(l[i])<m.get(l[j])){w=l[i];l[i]=l[j];l[j]=w;} for(;k<g;k++){x=76-l[k].length();y=m.get(l[k]);if(k<1||y*i>x*j){i=x;j=y;}} String s=new String(new char[m.get(l[0])*i/j]).replace('\0','_'); System.out.println(" "+s); for(k=0;k<g;k++){w=l[k];System.out.println("|"+s.substring(0,m.get(w)*i/j)+"| "+w);}} }

没有行为改进的版本是 615 字符：

import java.util.*;class F{public static void main(String[]l)throws Exception{Map<String,Integer>m=new HashMap();String w="";int i=0,k=0,j=8,g=22;for(;j>0;){j=System.in.read();if(j>90)j-=32;if(j>64&j<91)w+=(char)j;else{if(!w.matches("^(|.|THE|AND|OF|TO|I[TNS]|OR)$"))m.put(w,m.get(w)!=null?m.get(w)+1:1);w="";}}l=m.keySet().toArray(l);for(;i<g;++i)for(j=i;++j<l.length;)if(m.get(l[i])<m.get(l[j])){w=l[i];l[i]=l[j];l[j]=w;}i=76-l[0].length();String s=new String(new char[i]).replace('\0','_');System.out.println(" "+s);for(k=0;k<g;k++){w=l[k];System.out.println("|"+s.substring(0,m.get(w)*i/m.get(l[0]))+"| "+w);}}}

Answer 27

Scala 2.8,311 314 320 330 332 336 341 375 个字符

包括长字调整。借鉴其他解决方案的想法。

现在作为脚本（a.scala）：

val t="\\w+\\b(?<!\\bthe|and|of|to|a|i[tns]?|or)".r.findAllIn(io.Source.fromFile(argv(0)).mkString.toLowerCase).toSeq.groupBy(w=>w).mapValues(_.size).toSeq.sortBy(-_._2)take 22
def b(p:Int)="_"*(p*(for((w,c)<-t)yield(76.0-w.size)/c).min).toInt
println(" "+b(t(0)._2))
for(p<-t)printf("|%s| %s \n",b(p._2),p._1)

使用

运行

scala -howtorun:script a.scala alice.txt

顺便说一句，从314到311个字符的编辑实际上只删除了1个字符。之前有人弄错了（Windows CRs？）。

Answer 28

Clojure 282严格

(let[[[_ m]:as s](->>(slurp *in*).toLowerCase(re-seq #"\w+\b(?<!\bthe|and|of|to|a|i[tns]?|or)")frequencies(sort-by val >)(take 22))[b](sort(map #(/(- 76(count(key %)))(val %))s))p #(do(print %1)(dotimes[_(* b %2)](print \_))(apply println %&))](p " " m)(doseq[[k v]s](p \| v \| k)))

更清晰：

(let[[[_ m]:as s](->> (slurp *in*)
                   .toLowerCase
                   (re-seq #"\w+\b(?<!\bthe|and|of|to|a|i[tns]?|or)")
                   frequencies
                   (sort-by val >)
                   (take 22))
     [b] (sort (map #(/ (- 76 (count (key %)))(val %)) s))
     p #(do
          (print %1)
          (dotimes[_(* b %2)] (print \_))
          (apply println %&))]
  (p " " m)
  (doseq[[k v] s] (p \| v \| k)))

Answer 29

Scala，368个字符

首先，592个字符的清晰版本：

object Alice {
  def main(args:Array[String]) {
    val s = io.Source.fromFile(args(0))
    val words = s.getLines.flatMap("(?i)\\w+\\b(?<!\\bthe|and|of|to|a|i|it|in|or|is)".r.findAllIn(_)).map(_.toLowerCase)
    val freqs = words.foldLeft(Map[String, Int]())((countmap, word)  => countmap + (word -> (countmap.getOrElse(word, 0)+1)))
    val sortedFreqs = freqs.toList.sort((a, b)  => a._2 > b._2)
    val top22 = sortedFreqs.take(22)
    val highestWord = top22.head._1
    val highestCount = top22.head._2
    val widest = 76 - highestWord.length
    println(" " + "_" * widest)
    top22.foreach(t => {
      val width = Math.round((t._2 * 1.0 / highestCount) * widest).toInt
      println("|" + "_" * width + "| " + t._1)
    })
  }
}

控制台输出如下所示：

$ scalac alice.scala 
$ scala Alice aliceinwonderland.txt
 _________________________________________________________________________
|_________________________________________________________________________| she
|_______________________________________________________________| you
|_____________________________________________________________| said
|_____________________________________________________| alice
|_______________________________________________| was
|____________________________________________| that
|____________________________________| as
|_________________________________| her
|______________________________| at
|______________________________| with
|_____________________________| s
|_____________________________| t
|___________________________| on
|__________________________| all
|_______________________| had
|_______________________| but
|______________________| be
|______________________| not
|____________________| they
|____________________| so
|___________________| very
|___________________| what

我们可以做一些积极的缩小并将其降低到415个字符：

object A{def main(args:Array[String]){val l=io.Source.fromFile(args(0)).getLines.flatMap("(?i)\\w+\\b(?<!\\bthe|and|of|to|a|i|it|in|or|is)".r.findAllIn(_)).map(_.toLowerCase).foldLeft(Map[String, Int]())((c,w)=>c+(w->(c.getOrElse(w,0)+1))).toList.sort((a,b)=>a._2>b._2).take(22);println(" "+"_"*(76-l.head._1.length));l.foreach(t=>println("|"+"_"*Math.round((t._2*1.0/l.head._2)*(76-l.head._1.length)).toInt+"| "+t._1))}}

控制台会话如下所示：

$ scalac a.scala 
$ scala A aliceinwonderland.txt
 _________________________________________________________________________
|_________________________________________________________________________| she
|_______________________________________________________________| you
|_____________________________________________________________| said
|_____________________________________________________| alice
|_______________________________________________| was
|____________________________________________| that
|____________________________________| as
|_________________________________| her
|______________________________| at
|______________________________| with
|_____________________________| s
|_____________________________| t
|___________________________| on
|__________________________| all
|_______________________| had
|_______________________| but
|______________________| be
|______________________| not
|____________________| they
|____________________| so
|___________________| very
|___________________| what

我确信Scala专家可以做得更好。

更新：在评论中，托马斯提供了更短的版本，共368个字符：

object A{def main(a:Array[String]){val t=(Map[String, Int]()/:(for(x<-io.Source.fromFile(a(0)).getLines;y<-"(?i)\\w+\\b(?<!\\bthe|and|of|to|a|i|it|in|or|is)".r findAllIn x) yield y.toLowerCase).toList)((c,x)=>c+(x->(c.getOrElse(x,0)+1))).toList.sortBy(_._2).reverse.take(22);val w=76-t.head._1.length;print(" "+"_"*w);t map (s=>"\n|"+"_"*(s._2*w/t.head._2)+"| "+s._1) foreach print}}

清晰，375个字符：

object Alice {
  def main(a:Array[String]) {
    val t = (Map[String, Int]() /: (
      for (
        x <- io.Source.fromFile(a(0)).getLines
        y <- "(?i)\\w+\\b(?<!\\bthe|and|of|to|a|i|it|in|or|is)".r.findAllIn(x)
      ) yield y.toLowerCase
    ).toList)((c, x) => c + (x -> (c.getOrElse(x, 0) + 1))).toList.sortBy(_._2).reverse.take(22)
    val w = 76 - t.head._1.length
    print (" "+"_"*w)
    t.map(s => "\n|" + "_" * (s._2 * w / t.head._2) + "| " + s._1).foreach(print)
  }
}

Answer 30

C ++，647个字符

我不希望通过使用C ++获得高分，但是没关系。我很确定它符合所有要求。请注意，我使用C ++ 0x auto关键字进行变量声明，因此如果您决定测试我的代码，请相应地调整编译器。

最小化版本

#include <iostream>
#include <cstring>
#include <map>
using namespace std;
#define C string
#define S(x)v=F/a,cout<<#x<<C(v,'_')
#define F t->first
#define G t->second
#define O &&F!=
#define L for(i=22;i-->0;--t)
int main(){map<C,int>f;char d[230];int i=1,v;for(;i<256;i++)d[i<123?i-1:i-27]=i;d[229]=0;char w[99];while(cin>>w){for(i=0;w[i];i++)w[i]=tolower(w[i]);char*p=strtok(w,d);while(p)++f[p],p=strtok(0,d);}multimap<int,C>c;for(auto t=f.end();--t!=f.begin();)if(F!="the"O"and"O"of"O"to"O"a"O"i"O"it"O"in"O"or"O"is")c.insert(pair<int,C>(G,F));auto t=--c.end();float a=0,A;L A=F/(76.0-G.length()),a=a>A?a:A;t=--c.end();S( );L S(\n|)<<"| "<<G;}

这是使用string而不是char[]和strtok的更多“C ++”的第二个版本。它有点大，在 669（+22 vs above），但我现在不能让它变小，所以我认为我会发布它。

#include <iostream>
#include <map>
using namespace std;
#define C string
#define S(x)v=F/a,cout<<#x<<C(v,'_')
#define F t->first
#define G t->second
#define O &&F!=
#define L for(i=22;i-->0;--t)
#define E e=w.find_first_of(d,g);g=w.find_first_not_of(d,e);
int main(){map<C,int>f;int i,v;C w,x,d="abcdefghijklmnopqrstuvwxyz";while(cin>>w){for(i=w.size();i-->0;)w[i]=tolower(w[i]);unsigned g=0,E while(g-e>0){x=w.substr(e,g-e),++f[x],E}}multimap<int,C>c;for(auto t=f.end();--t!=f.begin();)if(F!="the"O"and"O"of"O"to"O"a"O"i"O"it"O"in"O"or"O"is")c.insert(pair<int,C>(G,F));auto t=--c.end();float a=0,A;L A=F/(76.0-G.length()),a=a>A?a:A;t=--c.end();S( );L S(\n|)<<"| "<<G;}

我删除了完整版本，因为我不会因为我对最小化版本的调整而不断更新它。如果您对（可能过时的）长版本感兴趣，请参阅编辑历史记录。

Answer 31

又一个python 2.x - 206个字符（或232个“宽度条”）

如果完全符合这个问题，我相信这个。忽略列表在这里，它完全检查行长度（参见我用Alice替换Aliceinwonderlandbylewiscarroll的例子，通过文本使第五项成为最长行。甚至文件名都是从命令行而不是硬编码提供的（硬编码它会删除大约10个字符。）它有一个缺点（但我相信问题还可以），因为它计算整数分频器使行短于80个字符，最长行短于80个字符，不完全是80个字符python 3.x版本没有这个缺陷（但是更长）。

此外，我认为阅读并不难。

import sys,re
t=re.split("\W+(?:(?:the|and|o[fr]|to|a|i[tns]?)\W+)*",sys.stdin.read().lower())
b=sorted((-t.count(x),x)for x in set(t))[:22]
for l,w in b:print"|"+l/min(z/(78-len(e))for z,e in b)*'-'+"|",w

|----------------------------------------------------------------| she
|--------------------------------------------------------| you
|-----------------------------------------------------| said
|----------------------------------------------| aliceinwonderlandbylewiscarroll
|-----------------------------------------| was
|--------------------------------------| that
|-------------------------------| as
|----------------------------| her
|--------------------------| at
|--------------------------| with
|-------------------------| s
|-------------------------| t
|-----------------------| on
|-----------------------| all
|---------------------| this
|--------------------| for
|--------------------| had
|--------------------| but
|-------------------| be
|-------------------| not
|------------------| they
|-----------------| so

目前尚不清楚我们是否必须单独打印最大条（如样本输出）。下面是另一个这样做，但232个字符。

import sys,re
t=re.split("\W+(?:(?:the|and|o[fr]|to|a|i[tns]?)\W+)*",sys.stdin.read().lower())
b=sorted((-t.count(x),x)for x in set(t))[:22]
f=min(z/(78-len(e))for z,e in b)
print"",b[0][0]/f*'-'
for y,w in b:print"|"+y/f*'-'+"|",w

Python 3.x - 256个字符

使用python 3.x中的Counter类，很有希望缩短它（因为Counter会执行我们需要的所有内容）。它出来并不是更好。以下是我的试用266个字符：

import sys,re,collections as c
b=c.Counter(re.split("\W+(?:(?:the|and|o[fr]|to|a|i[tns]?)\W+)*",
sys.stdin.read().lower())).most_common(22)
F=lambda p,x,w:print(p+'-'*int(x/max(z/(77.-len(e))for e,z in b))+w)
F(" ",b[0][1],"")
for w,y in b:F("|",y,"| "+w)

问题是，collections和most_common是非常长的单词，即使Counter也不短......真的，不使用Counter会使代码仅延长2个字符 - （

python 3.x还引入了其他约束：将两个整数分开不再是一个整数（所以我们必须转换为int），print现在是一个函数（必须添加括号）等等。这就是它出来的原因22字符长于python2.x版本，但速度更快。也许一些更实验的python 3.x编码器会有想法来缩短代码。

Answer 32

Java - 896个字符

931 chars

1233个字符不可读

1977 chars“uncompressed”

更新：我积极减少字符数。根据更新的规范省略单字母单词。

我非常羡慕C＃和LINQ。

import java.util.*;import java.io.*;import static java.util.regex.Pattern.*;class g{public static void main(String[] a)throws Exception{PrintStream o=System.out;Map<String,Integer> w=new HashMap();Scanner s=new Scanner(new File(a[0])).useDelimiter(compile("[^a-z]+|\\b(the|and|of|to|.|it|in|or|is)\\b",2));while(s.hasNext()){String z=s.next().trim().toLowerCase();if(z.equals(""))continue;w.put(z,(w.get(z)==null?0:w.get(z))+1);}List<Integer> v=new Vector(w.values());Collections.sort(v);List<String> q=new Vector();int i,m;i=m=v.size()-1;while(q.size()<22){for(String t:w.keySet())if(!q.contains(t)&&w.get(t).equals(v.get(i)))q.add(t);i--;}int r=80-q.get(0).length()-4;String l=String.format("%1$0"+r+"d",0).replace("0","_");o.println(" "+l);o.println("|"+l+"| "+q.get(0)+" ");for(i=m-1;i>m-22;i--){o.println("|"+l.substring(0,(int)Math.round(r*(v.get(i)*1.0)/v.get(m)))+"| "+q.get(m-i)+" ");}}}

“读”：

import java.util.*;
import java.io.*;
import static java.util.regex.Pattern.*;
class g
{
   public static void main(String[] a)throws Exception
      {
      PrintStream o = System.out;
      Map<String,Integer> w = new HashMap();
      Scanner s = new Scanner(new File(a[0]))
         .useDelimiter(compile("[^a-z]+|\\b(the|and|of|to|.|it|in|or|is)\\b",2));
      while(s.hasNext())
      {
         String z = s.next().trim().toLowerCase();
         if(z.equals(""))
            continue;
         w.put(z,(w.get(z) == null?0:w.get(z))+1);
      }
      List<Integer> v = new Vector(w.values());
      Collections.sort(v);
      List<String> q = new Vector();
      int i,m;
      i = m = v.size()-1;
      while(q.size()<22)
      {
         for(String t:w.keySet())
            if(!q.contains(t)&&w.get(t).equals(v.get(i)))
               q.add(t);
         i--;
      }
      int r = 80-q.get(0).length()-4;
      String l = String.format("%1$0"+r+"d",0).replace("0","_");
      o.println(" "+l);
      o.println("|"+l+"| "+q.get(0)+" ");
      for(i = m-1; i > m-22; i--)
      {
         o.println("|"+l.substring(0,(int)Math.round(r*(v.get(i)*1.0)/v.get(m)))+"| "+q.get(m-i)+" ");
      }
   }
}

Alice的输出：

 _________________________________________________________________________
|_________________________________________________________________________| she
|_______________________________________________________________| you
|_____________________________________________________________| said
|_____________________________________________________| alice
|_______________________________________________| was
|____________________________________________| that
|____________________________________| as
|_________________________________| her
|______________________________| with
|______________________________| at
|___________________________| on
|__________________________| all
|________________________| this
|________________________| for
|_______________________| had
|_______________________| but
|______________________| be
|______________________| not
|____________________| they
|____________________| so
|___________________| very
|___________________| what

堂吉诃德的输出（也来自古腾堡）：

 ________________________________________________________________________
|________________________________________________________________________| that
|________________________________________________________| he
|______________________________________________| for
|__________________________________________| his
|________________________________________| as
|__________________________________| with
|_________________________________| not
|_________________________________| was
|________________________________| him
|______________________________| be
|___________________________| don
|_________________________| my
|_________________________| this
|_________________________| all
|_________________________| they
|________________________| said
|_______________________| have
|_______________________| me
|______________________| on
|______________________| so
|_____________________| you
|_____________________| quixote

Answer 33

Clojure - 611个字符（未最小化）

我尝试在尽可能多的惯用Clojure中编写代码。我对draw-chart函数并不感到骄傲，但我想这段代码会说明Clojure的简洁性。

(ns word-freq
(:require [clojure.contrib.io :as io]))

(defn word-freq
  [f]
  (take 22 (->> f
                io/read-lines ;;; slurp should work too, but I love map/red
                (mapcat (fn [l] (map #(.toLowerCase %) (re-seq #"\w+" l))))
                (remove #{"the" "and" "of" "to" "a" "i" "it" "in" "or" "is"})
                (reduce #(assoc %1 %2 (inc (%1 %2 0))) {})
                (sort-by (comp - val)))))

(defn draw-chart
  [fs]
  (let [[[w f] & _] fs]
    (apply str
           (interpose \newline
                      (map (fn [[k v]] (apply str (concat "|" (repeat (int (* (- 76 (count w)) (/ v f 1))) "_") "| " k " ")) ) fs)))))

;;; (println (draw-chart (word-freq "/Users/ghoseb/Desktop/alice.txt")))

输出：

|_________________________________________________________________________| she 
|_______________________________________________________________| you 
|____________________________________________________________| said 
|____________________________________________________| alice 
|_______________________________________________| was 
|___________________________________________| that 
|____________________________________| as 
|________________________________| her 
|_____________________________| with 
|_____________________________| at 
|____________________________| t 
|____________________________| s 
|__________________________| on 
|__________________________| all 
|_______________________| for 
|_______________________| had 
|_______________________| this 
|_______________________| but 
|______________________| be 
|_____________________| not 
|____________________| they 
|____________________| so

我知道，这不遵循规范，但是，嘿，这是一些非常干净的Clojure代码已经很小了:)）

Answer 34

Groovy， 424 389 378 321个字符

将b=map.get(a)替换为b=map[a]，用matcher / iterator替换split？

def r,s,m=[:],n=0;def p={println it};def w={"_".multiply it};(new URL(this.args[0]).text.toLowerCase()=~/\b\w+\b/).each{s=it;if(!(s==~/(the|and|of|to|a|i[tns]?|or)/))m[s]=m[s]==null?1:m[s]+1};m.keySet().sort{a,b->m[b]<=>m[a]}.subList(0,22).each{k->if(n++<1){r=(m[k]/(76-k.length()));p" "+w(m[k]/r)};p"|"+w(m[k]/r)+"|"+k}

（作为groovy脚本执行，URL为cmd line arg。无需导入！）

此处可读版本：

def r,s,m=[:],n=0;
def p={println it};
def w={"_".multiply it};
(new URL(this.args[0]).text.toLowerCase()
        =~ /\b\w+\b/
        ).each{
        s=it;
        if (!(s ==~/(the|and|of|to|a|i[tns]?|or)/))
            m[s] = m[s] == null ? 1 : m[s] + 1
        };
    m.keySet()
        .sort{
            a,b -> m[b] <=> m[a]
        }
        .subList(0,22).each{
            k ->
                if( n++ < 1 ){
                    r=(m[k]/(76-k.length()));
                    p " " + w(m[k]/r)
                };
                p "|" + w(m[k]/r) + "|" + k
}

Answer 35

R 449 chars

可能会缩短......

bar <- function(w, l)
    {
    b <- rep("-", l)
    s <- rep(" ", l)
    cat(" ", b, "\n|", s, "| ", w, "\n ", b, "\n", sep="")
    }

f <- "alice.txt"
e <- c("the", "and", "of", "to", "a", "i", "it", "in", "or", "is", "")
w <- unlist(lapply(readLines(file(f)), strsplit, s=" "))
w <- tolower(w)
w <- unlist(lapply(w, gsub, pa="[^a-z]", r=""))
u <- unique(w[!w %in% e])
n <- unlist(lapply(u, function(x){length(w[w==x])}))
o <- rev(order(n))
n <- n[o]
m <- 77 - max(unlist(lapply(u[1:22], nchar)))
n <- floor(m*n/n[1])
u <- u[o]

for (i in 1:22)
    bar(u[i], n[i])

Answer 36

MATLAB 335 ~~404~~ ~~410字节~~ ~~357字节。~~ ~~390字节。~~ < / p>

更新的代码现在是335个字符而不是404个字符，并且似乎在两个示例中都很好。

原始消息（对于404个字符的代码）

这个版本有点长，然而，它会正确缩放条的长度如果有一个字太长，那么没有一列超过80.

因此，我的代码是357字节而没有重新缩放，而410则需要重新缩放。

A=textscan(fopen('11.txt'),'%s','delimiter',' 0123456789,.!?-_*^:;=+\\/(){}[]@&#$%~`|"''');
s=lower(A{1});s(cellfun('length', s)<2)=[];s(ismember(s,{'the','and','of','to','it','in','or','is'}))=[];
[w,~,i]=unique(s);N=hist(i,max(i)); [j,k]=sort(N,'descend'); b=k(1:22); n=cellfun('length',w(b));
q=80*N(b)'/N(k(1))+n; q=floor(q*78/max(q)-n); for i=1:22, fprintf('%s| %s\n',repmat('_',1,l(i)),w{k(i)});end

结果：

___________________________________________________________________________| she
_________________________________________________________________| you
______________________________________________________________| said
_______________________________________________________| alice
________________________________________________| was
____________________________________________| that
_____________________________________| as
_________________________________| her
______________________________| at
______________________________| with
____________________________| on
___________________________| all
_________________________| this
________________________| for
________________________| had
________________________| but
_______________________| be
_______________________| not
_____________________| they
____________________| so
___________________| very
___________________| what

例如，用“superlongstringofridiculous”替换Alice in Wonderland文本中“you”的所有实例，我的代码将正确地缩放结果：

____________________________________________________________________| she
_________________________________________________________| superlongstringstring
________________________________________________________| said
_________________________________________________| alice
____________________________________________| was
________________________________________| that
_________________________________| as
______________________________| her
___________________________| with
___________________________| at
_________________________| on
________________________| all
_____________________| this
_____________________| for
_____________________| had
_____________________| but
____________________| be
____________________| not
__________________| they
__________________| so
_________________| very
_________________| what

这是更新的代码写得更清晰：

A=textscan(fopen('t'),'%s','delimiter','':'@');
s=lower(A{1});
s(cellfun('length', s)<2|ismember(s,{'the','and','of','to','it','in','or','is'}))=[];
[w,~,i]=unique(s);
N=hist(i,max(i)); 
[j,k]=sort(N,'descend'); 
n=cellfun('length',w(k));
q=80*N(k)'/N(k(1))+n; 
q=floor(q*78/max(q)-n); 
for i=1:22, 
    fprintf('%s| %s\n',repmat('_',1,q(i)),w{k(i)});
end

Answer 37

Python 2.6， 273 269 267 266个字符。

（编辑：道具给ChristopheD寻找个性化建议）

import sys,re
t=re.findall('[a-z]+',"".join(sys.stdin).lower())
d=sorted((t.count(w),w)for w in set(t)-set("the and of to a i it in or is".split()))[:-23:-1]
r=min((78.-len(m[1]))/m[0]for m in d)
print'','_'*(int(d[0][0]*r-2))
for(a,b)in d:print"|"+"_"*(int(a*r-2))+"|",b

输出：

 _________________________________________________________________________
|_________________________________________________________________________| she
|_______________________________________________________________| you
|____________________________________________________________| said
|____________________________________________________| alice
|______________________________________________| was
|__________________________________________| that
|___________________________________| as
|_______________________________| her
|____________________________| with
|____________________________| at
|___________________________| s
|___________________________| t
|_________________________| on
|_________________________| all
|______________________| this
|______________________| for
|______________________| had
|_____________________| but
|____________________| be
|____________________| not
|___________________| they
|__________________| so

Answer 38

Java - 991 chars _{^{（包括换行符和缩进）}}

我拿了@seanizer的代码，修正了一个错误（他省略了第一个输出行），做了一些改进，使代码更“高尔夫”。

import java.util.*;
import java.util.regex.*;
import org.apache.commons.io.IOUtils;
public class WF{
 public static void main(String[] a)throws Exception{
  String t=IOUtils.toString(new java.net.URL(a[0]).openStream());
  class W implements Comparable<W> {
   String w;int f=1;W(String W){w=W;}public int compareTo(W o){return o.f-f;}
   String d(float r){char[]c=new char[(int)(f/r)];Arrays.fill(c,'_');return "|"+new String(c)+"| "+w;}
  }
  Map<String,W>M=new HashMap<String,W>();
  Matcher m=Pattern.compile("\\b\\w+\\b").matcher(t.toLowerCase());
  while(m.find()){String w=m.group();W W=M.get(w);if(W==null)M.put(w,new W(w));else W.f++;}
  M.keySet().removeAll(Arrays.asList("the,and,of,to,a,i,it,in,or,is".split(",")));
  List<W>L=new ArrayList<W>(M.values());Collections.sort(L);int l=76-L.get(0).w.length();
  System.out.println(" "+new String(new char[l]).replace('\0','_'));
  for(W w:L.subList(0,22))System.out.println(w.d((float)L.get(0).f/(float)l));
 }
}

输出：

 _________________________________________________________________________
|_________________________________________________________________________| she
|_______________________________________________________________| you
|____________________________________________________________| said
|_____________________________________________________| alice
|_______________________________________________| was
|___________________________________________| that
|____________________________________| as
|________________________________| her
|_____________________________| with
|_____________________________| at
|____________________________| s
|____________________________| t
|__________________________| on
|__________________________| all
|_______________________| this
|_______________________| for
|_______________________| had
|_______________________| but
|______________________| be
|_____________________| not
|____________________| they
|____________________| so

Answer 39

Shell，228个字符，80个字符约束工作

tr A-Z a-z|tr -Cs a-z "\n"|sort|egrep -v "^(the|and|of|to|a|i|it|in|or|is)$" |uniq -c|sort -r|head -22>g
n=1
while :
do
awk '{printf "|%0*s| %s\n",$1*'$n'/1e3,"",$2;}' g|tr 0 _>o 
egrep -q .{80} o&&break
n=$((n+1))
done
cat o

我很惊讶没有人似乎使用了printf的惊人*功能。

cat 11-very.txt＆gt; golf.sh

|__________________________________________________________________________| she
|________________________________________________________________| you
|_____________________________________________________________| said
|______________________________________________________| alice
|_______________________________________________| was
|____________________________________________| that
|____________________________________| as
|_________________________________| her
|______________________________| with
|______________________________| at
|_____________________________| s
|_____________________________| t
|___________________________| on
|__________________________| all
|________________________| this
|_______________________| for
|_______________________| had
|_______________________| but
|______________________| be
|______________________| not
|____________________| they
|____________________| so

猫11 | golf.sh

|_________________________________________________________________| she
|_________________________________________________________| verylongstringstring
|______________________________________________________| said
|_______________________________________________| alice
|__________________________________________| was
|_______________________________________| that
|________________________________| as
|_____________________________| her
|___________________________| with
|___________________________| at
|__________________________| s
|_________________________| t
|________________________| on
|_______________________| all
|_____________________| this
|_____________________| for
|_____________________| had
|____________________| but
|___________________| be
|___________________| not
|__________________| they
|__________________| so

Answer 40

Scala，327个字符

这是改编自mkneissl的answer灵感来自Python版本，虽然它更大。我将它留在这里以防有人可以缩短它。

val f="\\w+\\b(?<!\\bthe|and|of|to|a|i[tns]?|or)".r.findAllIn(io.Source.fromFile("11.txt").mkString.toLowerCase).toSeq
val t=f.toSet[String].map(x=> -f.count(x==)->x).toSeq.sorted take 22
def b(p:Int)="_"*(-p/(for((c,w)<-t)yield-c/(76.0-w.size)).max).toInt
println(" "+b(t(0)._1))
for(p<-t)printf("|%s| %s \n",b(p._1),p._2)

Answer 41

perl ，188个字符

上面的perl版本（以及任何基于regexp拆分的版本）可以通过将禁用词列表包含为负前瞻断言而不是单独的列表来缩短几个字节。此外，可以省略后面的分号。

我还提供了一些其他建议（ - 而不是＆lt; =＆gt;，for / foreach，drop“keys”）来到

$c{$_}++for grep{$_}map{lc=~/\b(?!(?:the|and|a|of|or|i[nts]?|to)\b)[a-z]+/g}<>;@s=sort{$c{$b}-$c{$a}}%c;$f=76-length$s[0];say$"."_"x$f;say"|"."_"x($c{$_}/$c{$s[0]}*$f)."| $_ "for@s[0..21]

我不知道perl，但我认为（？！（？：...）\ b）可能会丢失？：如果周围的处理是固定的。

Answer 42

TCL 554严格

foreach w [regexp -all -inline {[a-z]+} [string tolower [read stdin]]] {if {[lsearch {the and of to it in or is a i} $w]>=0} {continue};if {[catch {incr Ws($w)}]} {set Ws($w) 1}}
set T [lrange [lsort -decreasing -stride 2 -index 1 -integer [array get Ws]] 0 43]
foreach {w c} $T {lappend L [string length $w];lappend C $c}
set N [tcl::mathfunc::max {*}$L]
set C [lsort -integer $C]
set M [lindex $C end]
puts " [string repeat _ [expr {int((76-$N) * [lindex $T 1] / $M)}]] "
foreach {w c} $T {puts "|[string repeat _ [expr {int((76-$N) * $c / $M)}]]| $w"}

或者，更清晰

foreach w [regexp -all -inline {[a-z]+} [string tolower [read stdin]]] {
    if {[lsearch {the and of to a i it in or is} $w] >= 0} { continue }
    if {[catch {incr words($w)}]} {
        set words($w) 1
    }
}
set topwords [lrange [lsort -decreasing -stride 2 -index 1 -integer [array get words]] 0 43]
foreach {word count} $topwords {
    lappend lengths [string length $word]
    lappend counts $count
}
set maxlength [lindex [lsort -integer $lengths] end]
set counts [lsort -integer $counts]
set mincount [lindex $counts 0].0
set maxcount [lindex $counts end].0
puts " [string repeat _ [expr {int((76-$maxlength) * [lindex $topwords 1] / $maxcount)}]] "
foreach {word count} $topwords {
    set barlength [expr {int((76-$maxlength) * $count / $maxcount)}]
    puts "|[string repeat _ $barlength]| $word"
}

Answer 43

Go，613个字符，可能要小得多：

package main
import(r "regexp";. "bytes";. "io/ioutil";"os";st "strings";s "sort";. "container/vector")
type z struct{c int;w string}
func(e z)Less(o interface{})bool{return o.(z).c<e.c}
func main(){b,_:=ReadAll(os.Stdin);g:=r.MustCompile
c,m,x:=g("[A-Za-z]+").AllMatchesIter(b,0),map[string]int{},g("the|and|of|it|in|or|is|to")
for w:=range c{w=ToLower(w);if len(w)>1&&!x.Match(w){m[string(w)]++}}
o,y:=&Vector{},0
for k,v:=range m{o.Push(z{v,k});if v>y{y=v}}
s.Sort(o)
for i,v:=range *o{if i>21{break};x:=v.(z);c:=int(float(x.c)/float(y)*80)
u:=st.Repeat("_",c);if i<1{println(" "+u)};println("|"+u+"| "+x.w)}}

我觉得很脏。

Answer 44

Python 290 ， 255 ，253

python中的290个字符（从标准输入读取的文本）

import sys,re
c={}
for w in re.findall("[a-z]+",sys.stdin.read().lower()):c[w]=c.get(w,0)+1-(","+w+","in",a,i,the,and,of,to,it,in,or,is,")
r=sorted((-v,k)for k,v in c.items())[:22]
sf=max((76.0-len(k))/v for v,k in r)
print" "+"_"*int(r[0][0]*sf)
for v,k in r:print"|"+"_"*int(v*sf)+"| "+k

但......在阅读其他解决方案之后，我突然意识到效率不是一个要求;所以这是另一个更短，更慢的一个（255个字符）

import sys,re
w=re.findall("\w+",sys.stdin.read().lower())
r=sorted((-w.count(x),x)for x in set(w)-set("the and of to a i it in or is".split()))[:22]
f=max((76.-len(k))/v for v,k in r)
print" "+"_"*int(f*r[0][0])
for v,k in r:print"|"+"_"*int(f*v)+"| "+k

之后再阅读其他解决方案......

import sys,re
w=re.findall("\w+",sys.stdin.read().lower())
r=sorted((-w.count(x),x)for x in set(w)-set("the and of to a i it in or is".split()))[:22]
f=max((76.-len(k))/v for v,k in r)
print"","_"*int(f*r[0][0])
for v,k in r:print"|"+"_"*int(f*v)+"|",k

现在这个解决方案几乎每字节字节与Astatine的一个相同:-D

Answer 45

GNU Smalltalk（386）

我认为它可以缩短一点，但仍然不知道如何。

|q s f m|q:=Bag new. f:=FileStream stdin. m:=0.[f atEnd]whileFalse:[s:=f nextLine.(s notNil)ifTrue:[(s tokenize:'\W+')do:[:i|(((i size)>1)&({'the'.'and'.'of'.'to'.'it'.'in'.'or'.'is'}includes:i)not)ifTrue:[q add:(i asLowercase)]. m:=m max:(i size)]]].(q:=q sortedByCount)from:1to:22 do:[:i|'|'display.((i key)*(77-m)//(q first key))timesRepeat:['='display].('| %1'%{i value})displayNl]

Answer 46

另一个T-SQL解决方案借鉴了Martin's solution（min76-等）的一些想法。

declare @ varchar(max),@w real,@j int;select s=@ into[ ]set @=(select*
from openrowset(bulk'a',single_blob)a)while @>''begin set @=stuff(@,1,
patindex('%[a-z]%',@)-1,'')+'.'set @j=patindex('%[^a-z]%',@)if @j>2insert[ ]
select lower(left(@,@j-1))set @=stuff(@,1,@j,'')end;select top(22)s,count(*)
c into # from[ ]where',the,and,of,to,it,in,or,is,'not like'%,'+s+',%'
group by s order by 2desc;select @w=min((76.-len(s))/c),@=' '+replicate(
'_',max(c)*@w)from #;select @=@+'
|'+replicate('_',c*@w)+'| '+s+' 'from #;print @

整个解决方案应该在两行（先连接7），尽管你可以按原样剪切，粘贴和运行它。总字符数= 507 （如果以Unix格式保存并使用SQLCMD执行，则将换行符计为1）

假设：

没有临时表#
没有名为[ ]
输入位于默认系统文件夹中，例如C:\windows\system32\a
您的查询窗口'已设置nocount on'处于活动状态（防止虚假“受影响的行”msgs）

要进入解决方案列表（＆lt; 500 char），这里是 483 字符的“宽松”版本（没有竖条/没有顶栏/后面没有尾随空格）< / p>

declare @ varchar(max),@w real,@j int;select s=@ into[ ]set @=(select*
from openrowset(bulk'b',single_blob)a)while @>''begin set @=stuff(@,1,
patindex('%[a-z]%',@)-1,'')+'.'set @j=patindex('%[^a-z]%',@)if @j>2insert[ ]
select lower(left(@,@j-1))set @=stuff(@,1,@j,'')end;select top(22)s,count(*)
c into # from[ ]where',the,and,of,to,it,in,or,is,'not like'%,'+s+',%'
group by s order by 2desc;select @w=min((78.-len(s))/c),@=''from #;select @=@+'
'+replicate('_',c*@w)+' '+s from #;print @

可读版本

declare @ varchar(max), @w real, @j int
select s=@ into[ ] -- shortcut to create table; use defined variable to specify column type
-- openrowset reads an entire file
set @=(select * from openrowset(bulk'a',single_blob) a) -- a bit shorter than naming 'BulkColumn'

while @>'' begin -- loop until input is empty
    set @=stuff(@,1,patindex('%[a-z]%',@)-1,'')+'.' -- remove lead up to first A-Z char *
    set @j=patindex('%[^a-z]%',@) -- find first non A-Z char. The +'.' above makes sure there is one
    if @j>2insert[ ] select lower(left(@,@j-1)) -- insert only words >1 char
    set @=stuff(@,1,@j,'') -- remove word and trailing non A-Z char
end;

select top(22)s,count(*)c
into #
from[ ]
where ',the,and,of,to,it,in,or,is,' not like '%,'+s+',%' -- exclude list
group by s
order by 2desc; -- highest occurence, assume no ties at 22!

-- 80 - 2 vertical bars - 2 spaces = 76
-- @w = weighted frequency
-- this produces a line equal to the length of the max occurence (max(c))
select @w=min((76.-len(s))/c),@=' '+replicate('_',max(c)*@w)
from #;

-- for each word, append it as a new line. note: embedded newline
select @=@+'
|'+replicate('_',c*@w)+'| '+s+' 'from #;
-- note: 22 words in a table should always fit on an 8k page
--       the order of processing should always be the same as the insert-orderby
--       thereby producing the correct output

print @ -- output

Answer 47

Lua解决方案：478个字符。

t,u={},{}for l in io.lines()do
for w in l:gmatch("%a+")do
w=w:lower()if not(" the and of to a i it in or is "):find(" "..w.." ")then
t[w]=1+(t[w]or 0)end
end
end
for k,v in next,t do
u[#u+1]={k,v}end
table.sort(u,function(a,b)return a[2]>b[2]end)m,n=u[1][2],math.min(#u,22)for w=80,1,-1 do
s=""for i=1,n do
a,b=u[i][1],w*u[i][2]/m
if b+#a>=78 then s=nil break end
s2=("_"):rep(b)if i==1 then
s=s.." " ..s2.."\n"end
s=s.."|"..s2.."| "..a.."\n"end
if s then print(s)break end end

可读版本：

t,u={},{}
for line in io.lines() do
    for w in line:gmatch("%a+") do
        w = w:lower()
        if not (" the and of to a i it in or is "):find(" "..w.." ") then
            t[w] = 1 + (t[w] or 0)
        end
    end
end
for k, v in pairs(t) do
    u[#u+1]={k, v}
end

table.sort(u, function(a, b)
    return a[2] > b[2]
end)

local max = u[1][2]
local n = math.min(#u, 22)

for w = 80, 1, -1 do
    s=""
    for i = 1, n do
        f = u[i][2]
        word = u[i][1]
        width = w * f / max
        if width + #word >= 78 then
            s=nil
            break
        end
        s2=("_"):rep(width)
        if i==1 then
            s=s.." " .. s2 .."\n"
        end
        s=s.."|" .. s2 .. "| " .. word.."\n"
    end
    if s then
        print(s)
        break
    end
end

Answer 48

shell，grep，tr，grep，sort，uniq，sort，head，perl - 194 chars

添加一些-i标志可能会丢掉过长的tr A-Z a-z |步;规范没有说明所显示的案例，并且uniq -ci删除了任何案例差异。

egrep -oi [a-z]+|egrep -wiv 'the|and|o[fr]|to|a|i[tns]?'|sort|uniq -ci|sort -nr|head -22|perl -lape'($f,$w)=@F;$.>1or($q,$x)=($f,76-length$w);$b="_"x($f/$q*$x);$_="|$b| $w ";$.>1or$_=" $b\n$_"'

对于-i与原始的206个字符相比，对于tr加2，这是负11。

编辑：减去3的\\ b，可以省略，因为模式匹配无论如何都会在边界上开始。

sort首先给出小写，而uniq -ci首先出现，所以输出中唯一真正的变化就是Alice保留了她的大写首字母。

Answer 49

Bourne shell，213/240个字符

改进之前发布的shell版本，我可以将它降低到213个字符：

tr A-Z a-z|tr -Cs a-z \\n|sort|egrep -v '^(the|and|of|to|a|i|it|in|or|is)$'|uniq -c|sort -rn|sed 22q>g
n=1
>o
until egrep -q .{80} o
do
awk '{printf "|%0*d| %s\n",$1*'$n'/1e3,0,$2}' g|tr 0 _>o 
((n++))
done
cat o

为了在顶部栏上获得上部轮廓，我不得不将其扩展为240个字符：

tr A-Z a-z|tr -Cs a-z \\n|sort|egrep -v "^(the|and|of|to|a|i|it|in|or|is)$"|uniq -c|sort -r|sed 1p\;22q>g
n=1
>o
until egrep -q .{80} o
do
awk '{printf "|%0*d| %s\n",$1*'$n'/1e3,0,NR==1?"":$2}' g|sed '1s,|, ,g'|tr 0 _>o 
((n++))
done
cat o

Answer 50

Ruby，205

这个Ruby版本处理“superlongstringstring”。（前两行几乎与之前的Ruby程序相同。）

必须以这种方式运行：

ruby -n0777 golf.rb Alice.txt

W=($_.upcase.scan(/\w+/)-%w(THE AND OF TO A I IT
IN OR IS)).group_by{|x|x}.map{|k,v|[-v.size,k]}.sort[0,22]
u=proc{|m|"_"*(W.map{|n,s|(76.0-s.size)/n}.max*m)}
puts" "+u[W[0][0]],W.map{|n,s|"|%s| "%u[n]+s}

第三行创建一个闭包或lambda，产生一个正确缩放的下划线字符串：

u = proc{|m|
  "_" *
    (W.map{|n,s| (76.0 - s.size)/n}.max * m)
}

.max代替.min，因为数字是负数。

Answer 51

带有PC管道的对象Rexx 4.0

可以找到PC-Pipes库的位置。
这个解决方案忽略了单个字母的单词。


address rxpipe 'pipe (end ?) < Alice.txt',
   '|regex split /[^a-zA-Z]/', -- split at non alphbetic character
   '|locate 2',                -- discard words shorter that 2 char  
   '|xlate lower',             -- translate all words to lower case
   ,                           -- discard list words that match list
   '|regex not match /^(the||and||of||to||it||in||or||is)$/',
   '|l:lookup autoadd before count',  -- accumulate and count words
 '? l:',                       -- no master records to feed into lookup 
 '? l:',                       -- list of counted words comes here
   ,                           -- columns 1-10 hold count, 11-n hold word
   '|sort 1.10 d',             -- sort in desending order by count
   '|take 22',                 -- take first 22 records only
   '|array wordlist',          -- store into a rexx array
   '|count max',               -- get length of longest record 
   '|var maxword'              -- save into a rexx variable

parse value wordlist[1] with count 11 .  -- get frequency of first word
barunit = count % (76-(maxword-10))      -- frequency units per chart bar char

say ' '||copies('_', (count+barunit)%barunit)  -- first line of the chart
do cntwd over wordlist                    
  parse var cntwd count 11 word          -- get word frequency and the word
  say '|'||copies('_', (count+barunit)%barunit)||'| '||word||' '
end

输出产生了

 ________________________________________________________________________________
|________________________________________________________________________________| she
|_____________________________________________________________________| you
|___________________________________________________________________| said
|__________________________________________________________| alice
|____________________________________________________| was
|________________________________________________| that
|________________________________________| as
|____________________________________| her
|_________________________________| at
|_________________________________| with
|______________________________| on
|_____________________________| all
|__________________________| this
|__________________________| for
|__________________________| had
|__________________________| but
|________________________| be
|________________________| not
|_______________________| they
|______________________| so
|_____________________| very
|_____________________| what

Answer 52

喜欢大的...... Objective-C（ ~~1070~~ ~~931~~ 905 chars）

#define S NSString
#define C countForObject
#define O objectAtIndex
#define U stringWithCString
main(int g,char**b){id c=[NSCountedSet set];S*d=[S stringWithContentsOfFile:[S U:b[1]]];id p=[NSPredicate predicateWithFormat:@"SELF MATCHES[cd]'(the|and|of|to|a|i[tns]?|or)|[^a-z]'"];[d enumerateSubstringsInRange:NSMakeRange(0,[d length])options:NSStringEnumerationByWords usingBlock:^(S*s,NSRange x,NSRange y,BOOL*z){if(![p evaluateWithObject:s])[c addObject:[s lowercaseString]];}];id s=[[c allObjects]sortedArrayUsingComparator:^(id a,id b){return(NSComparisonResult)([c C:b]-[c C:a]);}];g=[c C:[s O:0]];int j=76-[[s O:0]length];char*k=malloc(80);memset(k,'_',80);S*l=[S U:k length:80];printf(" %s\n",[[l substringToIndex:j]cString]),[[s subarrayWithRange:NSMakeRange(0,22)]enumerateObjectsUsingBlock:^(id a,NSUInteger x,BOOL*y){printf("|%s| %s\n",[[l substringToIndex:[c C:a]*j/g]cString],[a cString]);}];}

切换到使用大量折旧API，删除了一些不需要的内存管理，更积极的空白删除

 _________________________________________________________________________
|_________________________________________________________________________| she
|______________________________________________________________| said
|__________________________________________________________| you
|____________________________________________________| alice
|________________________________________________| was
|_______________________________________| that
|____________________________________| as
|_________________________________| her
|______________________________| with
|______________________________| at
|___________________________| on
|__________________________| all
|________________________| this
|________________________| for
|________________________| had
|_______________________| but
|______________________| be
|______________________| not
|____________________| so
|___________________| very
|__________________| what
|_________________| they

Answer 53

R，298个字符

f=scan("stdin","ch")
u=unlist
s=strsplit
a=u(s(u(s(tolower(f),"[^a-z]")),"^(the|and|of|to|it|in|or|is|.|)$"))
v=unique(a)
r=sort(sapply(v,function(i) sum(a==i)),T)[2:23]  #the first item is an empty string, just skipping it
w=names(r)
q=(78-max(nchar(w)))*r/max(r)
cat(" ",rep("_",q[1])," \n",sep="")
for(i in 1:22){cat("|",rep("_",q[i]),"| ",w[i],"\n",sep="")}

输出结果为：

 _________________________________________________________________________ 
|_________________________________________________________________________| she
|_______________________________________________________________| you
|____________________________________________________________| said
|_____________________________________________________| alice
|_______________________________________________| was
|___________________________________________| that
|____________________________________| as
|________________________________| her
|_____________________________| at
|_____________________________| with
|__________________________| on
|__________________________| all
|_______________________| this
|_______________________| for
|_______________________| had
|_______________________| but
|______________________| be
|_____________________| not
|____________________| they
|____________________| so
|___________________| very
|__________________| what

如果“你”被更长的东西取代：

 ____________________________________________________________ 
|____________________________________________________________| she
|____________________________________________________| veryverylongstring
|__________________________________________________| said
|___________________________________________| alice
|______________________________________| was
|___________________________________| that
|_____________________________| as
|__________________________| her
|________________________| at
|________________________| with
|______________________| on
|_____________________| all
|___________________| this
|___________________| for
|___________________| had
|__________________| but
|__________________| be
|__________________| not
|________________| they
|________________| so
|_______________| very
|_______________| what

Answer 54

Python，320个字符

import sys
i="the and of to a i it in or is".split()
d={}
for j in filter(lambda x:x not in i,sys.stdin.read().lower().split()):d[j]=d.get(j,0)+1
w=sorted(d.items(),key=lambda x:x[1])[:-23:-1]
m=sorted(dict(w).values())[-1]
print" %s\n"%("_"*(76-m)),"\n".join(map(lambda x:("|%s| "+x[0])%("_"*((76-m)*x[1]/w[0][1])),w))

Answer 55

Java，慢慢变短（ ~~1500~~ ~~1358~~ ~~1241~~ ~~1020~~ ~~913~~ 890字符）

剥去更多的空格和var名称长度。尽可能删除泛型，删除内联类和try / catch块太糟糕了，我的900版有一个bug

删除了另一个try / catch块

import java.net.*;import java.util.*;import java.util.regex.*;import org.apache.commons.io.*;public class G{public static void main(String[]a)throws Exception{String text=IOUtils.toString(new URL(a[0]).openStream()).toLowerCase().replaceAll("\\b(the|and|of|to|a|i[tns]?|or)\\b","");final Map<String,Integer>p=new HashMap();Matcher m=Pattern.compile("\\b\\w+\\b").matcher(text);Integer b;while(m.find()){String w=m.group();b=p.get(w);p.put(w,b==null?1:b+1);}List<String>v=new Vector(p.keySet());Collections.sort(v,new Comparator(){public int compare(Object l,Object m){return p.get(m)-p.get(l);}});boolean t=true;float r=0;for(String w:v.subList(0,22)){if(t){t=false;r=p.get(w)/(float)(80-(w.length()+4));System.out.println(" "+new String(new char[(int)(p.get(w)/r)]).replace('\0','_'));}System.out.println("|"+new String(new char[(int)(((Integer)p.get(w))/r)]).replace('\0','_')+"|"+w);}}}

可读版本：

import java.net.*;
import java.util.*;
import java.util.regex.*;
import org.apache.commons.io.*;

public class G{

    public static void main(String[] a) throws Exception{
        String text =
            IOUtils.toString(new URL(a[0]).openStream())
                .toLowerCase()
                .replaceAll("\\b(the|and|of|to|a|i[tns]?|or)\\b", "");
        final Map<String, Integer> p = new HashMap();
        Matcher m = Pattern.compile("\\b\\w+\\b").matcher(text);
        Integer b;
        while(m.find()){
            String w = m.group();
            b = p.get(w);
            p.put(w, b == null ? 1 : b + 1);
        }
        List<String> v = new Vector(p.keySet());
        Collections.sort(v, new Comparator(){

            public int compare(Object l, Object m){
                return p.get(m) - p.get(l);
            }
        });
        boolean t = true;
        float r = 0;
        for(String w : v.subList(0, 22)){
            if(t){
                t = false;
                r = p.get(w) / (float) (80 - (w.length() + 4));
                System.out.println(" "
                    + new String(new char[(int) (p.get(w) / r)]).replace('\0',
                        '_'));
            }
            System.out.println("|"
                + new String(new char[(int) (((Integer) p.get(w)) / r)]).replace('\0',
                    '_') + "|" + w);
        }
    }
}

Answer 56

Javascript，348个字符

在我完成了我之后，我从Matt那里偷了一些想法：3

t=prompt().toLowerCase().replace(/\b(the|and|of|to|a|i[tns]?|or)\b/gm,'');r={};o=[];t.replace(/\b([a-z]+)\b/gm,function(a,w){r[w]?++r[w]:r[w]=1});for(i in r){o.push([i,r[i]])}m=o[0][1];o=o.slice(0,22);o.sort(function(F,D){return D[1]-F[1]});for(B in o){F=o[B];L=new Array(~~(F[1]/m*(76-F[0].length))).join('_');print(' '+L+'\n|'+L+'| '+F[0]+' \n')}

需要打印和提示功能支持。

Answer 57

Python， 250个字符

借用所有其他Python代码段

import re,sys
t=re.findall("\w+","".join(sys.stdin).lower())
W=sorted((-t.count(w),w)for w in set(t)-set("the and of to a i it in or is".split()))[:22]
Z,U=W[0],lambda n:"_"*int(n*(76.-len(Z[1]))/Z[0])
print"",U(Z[0])
for(n,w)in W:print"|"+U(n)+"|",w

如果你是厚颜无耻的并且要避免作为参数， 223 chars

import re,sys
t=re.findall("\w+","".join(sys.stdin).lower())
W=sorted((-t.count(w),w)for w in set(t)-set(sys.argv[1:]))[:22]
Z,U=W[0],lambda n:"_"*int(n*(76.-len(Z[1]))/Z[0])
print"",U(Z[0])
for(n,w)in W:print"|"+U(n)+"|",w

输出是：

$ python alice4.py  the and of to a i it in or is < 11.txt 
 _________________________________________________________________________
|_________________________________________________________________________| she
|_______________________________________________________________| you
|____________________________________________________________| said
|_____________________________________________________| alice
|_______________________________________________| was
|___________________________________________| that
|____________________________________| as
|________________________________| her
|_____________________________| at
|_____________________________| with
|____________________________| s
|____________________________| t
|__________________________| on
|__________________________| all
|_______________________| this
|_______________________| for
|_______________________| had
|_______________________| but
|______________________| be
|_____________________| not
|____________________| they
|____________________| so

Answer 58

Groovy，250

代码：

m=[:]
(new URL(args[0]).text.toLowerCase()=~/\w+/).each{it==~/(the|and|of|to|a|i[tns]?|or)/?:(m[it]=1+(m[it]?:0))}
k=m.keySet().sort{a,b->m[b]<=>m[a]}
b={d,c,b->println d+'_'*c+d+' '+b}
b' ',z=77-k[0].size(),''
k[0..21].each{b'|',m[it]*z/m[k[0]],it}

执行：

$ groovy wordcount.groovy http://www.gutenberg.org/files/11/11.txt

输出：

 __________________________________________________________________________  
|__________________________________________________________________________| she
|________________________________________________________________| you
|_____________________________________________________________| said
|_____________________________________________________| alice
|_______________________________________________| was
|____________________________________________| that
|____________________________________| as
|_________________________________| her
|______________________________| at
|______________________________| with
|_____________________________| s
|_____________________________| t
|___________________________| on
|__________________________| all
|________________________| this
|_______________________| for
|_______________________| had
|_______________________| but
|______________________| be
|______________________| not
|____________________| they
|____________________| so

N.B。这遵循宽松的规则：长串

Answer 59

Q，194

{t::y;{(-1')t#(.:)[b],'(!:)[b:"|",/:(((_)70*x%(*:)x)#\:"_"),\:"|"];}desc(#:')(=)($)(`$inter\:[(,/)" "vs'" "sv/:"'"vs'a(&)0<(#:')a:(_:')read0 -1!x;52#.Q.an])except`the`and`of`to`a`i`it`in`or`is`}

该函数有两个参数：一个是包含文本的文件，另一个是要显示的图表的行数

q){t::y;{(-1')t#(.:)[b],'(!:)[b:"|",/:(((_)70*x%(*:)x)#\:"_"),\:"|"];}desc(#:')(=)($)(`$inter\:[(,/)" "vs'" "sv/:"'"vs'a(&)0<(#:')a:(_:')read0 -1!x;52#.Q.an])except`the`and`of`to`a`i`it`in`or`is`}[`a.txt;20]

输出

|______________________________________________________________________|she
|____________________________________________________________|you
|__________________________________________________________|said
|___________________________________________________|alice
|_____________________________________________|was
|_________________________________________|that
|__________________________________|as
|_______________________________|her
|_____________________________|with
|____________________________|at
|___________________________|t
|___________________________|s
|_________________________|on
|_________________________|all
|_______________________|this
|______________________|for
|______________________|had
|_____________________|but
|_____________________|be
|_____________________|not

构建给定文本中最常用单词的ASCII图表

挑战：

59 个答案:

LabVIEW 51节点，5个结构，10个图

Ruby 1.9,185个字符

GolfScript， 177 175 173 167 164 163 < / s> 144 131 130个字符

基于Transact SQL set的解决方案（SQL Server 2005） 1063 892 873 853 827 < / s> 820 783 683 647 644 630个字符

206

shell，grep，tr，grep，sort，uniq，sort，head，perl

Ruby 207 213 211 210 207 203 <罢工> 201 200个字符

Mathematica（ 297 284 248 244 242 199 chars）Pure Functional

和Zipf的法律测试

Zipf的法律测试

编辑6＆gt; （242 Chars）

编辑7→199个字符

C＃ - 510 451 436 446 434 426 < / strike> 422 chars（minified）

Perl， 237 229 209 chars

Windows PowerShell，199个字符

Ruby，215， 216 ， 218 ， 221 ， 224 ， 236 ， 237 chars

Python 2.x，自由主义方法= 227 183 chars

咆哮

Python 2.x，punctilious approach = 277 243个字符

Haskell - 366 351 344 337 333个字符

Python 3.1 - 245 229个字符

PHP CLI版本（450个字符）

perl， 205 191 189个字符/ 205个字符（完全实现）

JavaScript 1.8（SpiderMonkey） - 354

JavaScript（Rhino） - 405 395 387 377 368 343 304个字符

Perl： 203 202 201 198 195 208 203/231 chars

F＃，452个字符

Python 2.6,347字符

通用LISP，670个字符

Gawk - 336（最初为507）个字符

* sh（+ curl），部分解决方案

Perl，185 char

C（828）

Java - 886 865 756 744 742 744 < / strike> 752 742 714 680个字符

Scala 2.8,311 314 320 330 332 336 341 375 个字符

Clojure 282严格

Scala，368个字符

C ++，647个字符

又一个python 2.x - 206个字符（或232个“宽度条”）

Python 3.x - 256个字符

Java - 896个字符

931 chars

1233个字符不可读

1977 chars“uncompressed”

Clojure - 611个字符（未最小化）

Groovy， 424 389 378 321个字符

R 449 chars

Python 2.6， 273 269 267 266个字符。

Java - 991 chars （包括换行符和缩进）

Shell，228个字符，80个字符约束工作

TCL 554严格

Python 290 ， 255 ，253

GNU Smalltalk（386）

可读版本

shell，grep，tr，grep，sort，uniq，sort，head，perl - 194 chars

Bourne shell，213/240个字符

Ruby，205

带有PC管道的对象Rexx 4.0

R，298个字符

Python，320个字符

删除了另一个try / catch块

Javascript，348个字符

Groovy，250

Q，194

Java - 991 chars _{^{（包括换行符和缩进）}}