Question

给出一个单词W，我想从/ usr / dict / words中找到包含W中字母的所有单词。例如，“bat”应该返回“bat”和“tab”（但不是“table”）。

这是一个解决方案，它涉及对输入词进行排序和匹配：

word=$1
sortedWord=`echo $word | grep -o . | sort | tr -d '\n'`

while read line
do
    sortedLine=`echo $line | grep -o . | sort | tr -d '\n'`
    if [ "$sortedWord" == "$sortedLine" ]
    then
        echo $line
    fi
done < /usr/dict/words

有更好的方法吗？我更喜欢使用基本命令（而不是perl / awk等），但欢迎所有解决方案！

为了澄清，我想找到原始单词的所有排列。不允许添加或删除字符。

Answer 1

这是一个awk实现。它在“W”中找到带有这些字母的单词。

dict="/usr/share/dict/words"
word=$1
awk -vw="$word" 'BEGIN{
  m=split(w,c,"")
  for(p=1;p<=m;p++){ chars[c[p]]++ }
}
length($0)==length(w){
  f=0;g=0
  n=split($0,t,"")
  for(o=1;o<=n;o++){
    if (!( t[o] in chars) ){
       f=1; break
    }else{ st[t[o]]++ }
  }
  if (!f || $0==w){
      for(z in st){
        if ( st[z] != chars[z] ) { g=1 ;break}
      }
      if(!g){ print "found: "$0 }
  }
  delete st
}' $dict

输出

$ wc -l < /usr/share/dict/words
479829

$ time ./shell.sh look
found: kolo
found: look

real    0m1.361s
user    0m1.074s
sys     0m0.015s

更新：使用排序

更改算法

dict="/usr/share/dict/words"
awk 'BEGIN{
  w="table"
  m=split(w,c,"")
  b=asort(c,chars)
}
length($0)==length(w){
  f=0
  n=split($0,t,"")
  e=asort(t,d)
  for(i=1;i<=e;i++) {
    if(d[i]!=chars[i]){
        f=1;break
    }
  }
  if(!f) print $0
}' $dict

输出

$ time ./shell.sh #looking for table
ablet
batel
belat
blate
bleat
tabel
table

real    0m1.416s
user    0m1.343s
sys     0m0.014s

$ time ./shell.sh #looking for chairs
chairs
ischar
rachis

real    0m1.697s
user    0m1.660s
sys     0m0.014s

$ time perl perl.pl #using beamrider's Perl script
table
tabel
ablet
batel
blate
bleat
belat

real    0m2.680s
user    0m1.633s
sys     0m0.881s

$ time perl perl.pl # looking for chairs
chairs
ischar
rachis

real    0m14.044s
user    0m8.328s
sys     0m5.236s

Answer 2

这是一个shell解决方案。最好的算法似乎是＃4。它会过滤掉所有长度不正确的单词。然后，它使用简单的替换密码（a = 1，b = 2，A = 27，...）对单词求和。如果总和匹配，那么它实际上将进行原始排序并进行比较。在我的系统上，它可以在不到1/2秒的时间内通过~235k的单词来寻找“蝙蝠”。我提供了所有解决方案，因此您可以看到不同的方法。

更新：未显示，但我也尝试将总和放在我尝试的直方图方法的第一个bin中，但它甚至比没有的直方图更慢。我认为它会起到短路的作用，但它不起作用。

Update2：我尝试了awk解决方案，它运行时间约为我最佳shell解决方案的1/3或~0.126s对比~0.490s。 perl解决方案运行~1.1s。

#!/bin/bash

word=$1
#dict=words
dict=/usr/share/dict/words
#dict=/usr/dict/words

alg1() {
  sortedWord=`echo $word | grep -o . | sort | tr -d '\n'`

  while read line
  do
    sortedLine=`echo $line | grep -o . | sort | tr -d '\n'`
    if [ "$sortedWord" == "$sortedLine" ]
    then
      echo $line
    fi
  done < $dict
}

check_sorted_versus_not() {
    local word=$1
    local line=`echo $2 | grep -o . | sort | tr -d '\n'`
    if [ "$word" == "$line" ]
    then
        echo $2
    fi
}

# Filter out all words of incorrect length
alg2() {
  sortedWord=`echo $word | grep -o . | sort | tr -d '\n'`
  grep_string="^`echo -n $word | tr 'a-zA-Z' '.'`\$"

  grep "$grep_string" "$dict" | \
  while read line
  do
    sortedLine=`echo $line | grep -o . | sort | tr -d '\n'`
    if [ "$sortedWord" == "$sortedLine" ]
    then
      echo $line
    fi
  done
}


# Create a lot of variables like this:
# _a=1, _b=2, ... _z=26, _A=27, _B=28, ... _Z=52
gen_chars() {
#  [ -n "$GEN_CHARS" ] && return
  GEN_CHARS=1
  local alpha="abcdefghijklmnopqrstuvwxyz"
  local upperalpha=`echo -n $alpha | tr 'a-z' 'A-Z'`
  local both="$alpha$upperalpha"
  for ((i=0; i < ${#both}; i++))
  do
    ACHAR=${both:i:1}
    eval "_$ACHAR=$((i+1))"
  done
}

# I think it's faster to return the value in a var then to echo it in a sub process.
# Try summing the word one char at a time by building an arithmetic expression
# and then evaluate that expression.
# Requires: gen_chars
sum_word() {
  SUM=0
  local s=""
  # parsing input one character at a time
  for ((i=0; i < ${#1}; i++))
  do
    ACHAR=${1:i:1}
    s="$s\$_$ACHAR+"
  done

  SUM=$(( $(eval echo -n ${s}0) ))
}

# I think it's faster to return the value in a var then to echo it in a sub process.
# Try summing the word one char at a time using a case statement.
sum_word2() {
  SUM=0
  local s=""
  # parsing input one character at a time
  for ((i=0; i < ${#1}; i++))
  do
    ACHAR=${1:i:1}
    case $ACHAR in
    a) SUM=$((SUM+  1));;
    b) SUM=$((SUM+  2));;
    c) SUM=$((SUM+  3));;
    d) SUM=$((SUM+  4));;
    e) SUM=$((SUM+  5));;
    f) SUM=$((SUM+  6));;
    g) SUM=$((SUM+  7));;
    h) SUM=$((SUM+  8));;
    i) SUM=$((SUM+  9));;
    j) SUM=$((SUM+ 10));;
    k) SUM=$((SUM+ 11));;
    l) SUM=$((SUM+ 12));;
    m) SUM=$((SUM+ 13));;
    n) SUM=$((SUM+ 14));;
    o) SUM=$((SUM+ 15));;
    p) SUM=$((SUM+ 16));;
    q) SUM=$((SUM+ 17));;
    r) SUM=$((SUM+ 18));;
    s) SUM=$((SUM+ 19));;
    t) SUM=$((SUM+ 20));;
    u) SUM=$((SUM+ 21));;
    v) SUM=$((SUM+ 22));;
    w) SUM=$((SUM+ 23));;
    x) SUM=$((SUM+ 24));;
    y) SUM=$((SUM+ 25));;
    z) SUM=$((SUM+ 26));;
    A) SUM=$((SUM+ 27));;
    B) SUM=$((SUM+ 28));;
    C) SUM=$((SUM+ 29));;
    D) SUM=$((SUM+ 30));;
    E) SUM=$((SUM+ 31));;
    F) SUM=$((SUM+ 32));;
    G) SUM=$((SUM+ 33));;
    H) SUM=$((SUM+ 34));;
    I) SUM=$((SUM+ 35));;
    J) SUM=$((SUM+ 36));;
    K) SUM=$((SUM+ 37));;
    L) SUM=$((SUM+ 38));;
    M) SUM=$((SUM+ 39));;
    N) SUM=$((SUM+ 40));;
    O) SUM=$((SUM+ 41));;
    P) SUM=$((SUM+ 42));;
    Q) SUM=$((SUM+ 43));;
    R) SUM=$((SUM+ 44));;
    S) SUM=$((SUM+ 45));;
    T) SUM=$((SUM+ 46));;
    U) SUM=$((SUM+ 47));;
    V) SUM=$((SUM+ 48));;
    W) SUM=$((SUM+ 49));;
    X) SUM=$((SUM+ 50));;
    Y) SUM=$((SUM+ 51));;
    Z) SUM=$((SUM+ 52));;
    *) SUM=0; return;;
    esac
  done
}

# I think it's faster to return the value in a var then to echo it in a sub process.
# Try summing the word by building an arithmetic expression using sed and then evaluating
# the expression.
# Requires: gen_chars
sum_word3() {
  SUM=$(( $(eval echo -n `echo -n $1 | sed -E -ne 's,.,$_&+,pg'`) 0))
  #echo "SUM($1)=$SUM"
}

# Filter out all words of incorrect length
# Sum the characters in the word: i.e. a=1, b=2, ...  and "abbc" = 1+2+2+3 = 8
alg3() {
  gen_chars
  sortedWord=`echo $word | grep -o . | sort | tr -d '\n'`
  sum_word $word
  word_sum=$SUM
  grep_string="^`echo -n $word | tr 'a-zA-Z' '.'`\$"

  grep "$grep_string" "$dict" | \
  while read line
  do
    sum_word $line
    line_sum=$SUM
    if [ $word_sum == $line_sum ]
    then
      check_sorted_versus_not $sortedWord $line
    fi
  done
}

# Filter out all words of incorrect length
# Sum the characters in the word: i.e. a=1, b=2, ...  and "abbc" = 1+2+2+3 = 8
# Use sum_word2
alg4() {
  sortedWord=`echo $word | grep -o . | sort | tr -d '\n'`
  sum_word2 $word
  word_sum=$SUM
  grep_string="^`echo -n $word | tr 'a-zA-Z' '.'`\$"

  grep "$grep_string" "$dict" | \
  while read line
  do
    sum_word2 $line
    line_sum=$SUM
    if [ $word_sum == $line_sum ]
    then
      check_sorted_versus_not $sortedWord $line
    fi
  done
}

# Filter out all words of incorrect length
# Sum the characters in the word: i.e. a=1, b=2, ...  and "abbc" = 1+2+2+3 = 8
# Use sum_word3
alg5() {
  gen_chars
  sortedWord=`echo $word | grep -o . | sort | tr -d '\n'`
  sum_word3 $word
  word_sum=$SUM
  grep_string="^`echo -n $word | tr 'a-zA-Z' '.'`\$"

  grep "$grep_string" "$dict" | \
  while read line
  do
    sum_word3 $line
    line_sum=$SUM
    if [ $word_sum == $line_sum ]
    then
      check_sorted_versus_not $sortedWord $line
    fi
  done
}


# I think it's faster to return the value in a var then to echo it in a sub process.
# Try summing the word one char at a time using a case statement.
# Place results in a histogram
sum_word4() {
  SUM=(0 0 0 0 0 0 0 0 0 0
       0 0 0 0 0 0 0 0 0 0
       0 0 0 0 0 0 
       0 0 0 0 0 0 0 0 0 0
       0 0 0 0 0 0 0 0 0 0
       0 0 0 0 0 0 
       0)
  # parsing input one character at a time
  for ((i=0; i < ${#1}; i++))
  do
    ACHAR=${1:i:1}
    case $ACHAR in
    a) SUM[1]=$((SUM[ 1] + 1));;
    b) SUM[2]=$((SUM[ 2] + 1));;
    c) SUM[3]=$((SUM[ 3] + 1));;
    d) SUM[4]=$((SUM[ 4] + 1));;
    e) SUM[5]=$((SUM[ 5] + 1));;
    f) SUM[6]=$((SUM[ 6] + 1));;
    g) SUM[7]=$((SUM[ 7] + 1));;
    h) SUM[8]=$((SUM[ 8] + 1));;
    i) SUM[9]=$((SUM[ 9] + 1));;
    j) SUM[10]=$((SUM[10] + 1));;
    k) SUM[11]=$((SUM[11] + 1));;
    l) SUM[12]=$((SUM[12] + 1));;
    m) SUM[13]=$((SUM[13] + 1));;
    n) SUM[14]=$((SUM[14] + 1));;
    o) SUM[15]=$((SUM[15] + 1));;
    p) SUM[16]=$((SUM[16] + 1));;
    q) SUM[17]=$((SUM[17] + 1));;
    r) SUM[18]=$((SUM[18] + 1));;
    s) SUM[19]=$((SUM[19] + 1));;
    t) SUM[20]=$((SUM[20] + 1));;
    u) SUM[21]=$((SUM[21] + 1));;
    v) SUM[22]=$((SUM[22] + 1));;
    w) SUM[23]=$((SUM[23] + 1));;
    x) SUM[24]=$((SUM[24] + 1));;
    y) SUM[25]=$((SUM[25] + 1));;
    z) SUM[26]=$((SUM[26] + 1));;
    A) SUM[27]=$((SUM[27] + 1));;
    B) SUM[28]=$((SUM[28] + 1));;
    C) SUM[29]=$((SUM[29] + 1));;
    D) SUM[30]=$((SUM[30] + 1));;
    E) SUM[31]=$((SUM[31] + 1));;
    F) SUM[32]=$((SUM[32] + 1));;
    G) SUM[33]=$((SUM[33] + 1));;
    H) SUM[34]=$((SUM[34] + 1));;
    I) SUM[35]=$((SUM[35] + 1));;
    J) SUM[36]=$((SUM[36] + 1));;
    K) SUM[37]=$((SUM[37] + 1));;
    L) SUM[38]=$((SUM[38] + 1));;
    M) SUM[39]=$((SUM[39] + 1));;
    N) SUM[40]=$((SUM[40] + 1));;
    O) SUM[41]=$((SUM[41] + 1));;
    P) SUM[42]=$((SUM[42] + 1));;
    Q) SUM[43]=$((SUM[43] + 1));;
    R) SUM[44]=$((SUM[44] + 1));;
    S) SUM[45]=$((SUM[45] + 1));;
    T) SUM[46]=$((SUM[46] + 1));;
    U) SUM[47]=$((SUM[47] + 1));;
    V) SUM[48]=$((SUM[48] + 1));;
    W) SUM[49]=$((SUM[49] + 1));;
    X) SUM[50]=$((SUM[50] + 1));;
    Y) SUM[51]=$((SUM[51] + 1));;
    Z) SUM[52]=$((SUM[52] + 1));;
    *) SUM[53]=-1; return;;
    esac
  done

 #echo ${SUM[*]}
}

# Check if two histograms are equal
hist_are_equal() {
  # Array sizes differ?
  [ ${#_h1[*]} != ${#SUM[*]} ] && return 1

  # parsing input one index at a time
  for ((i=0; i < ${#_h1[*]}; i++))
  do
    [ ${_h1[i]} != ${SUM[i]} ] && return 1
  done

  return 0
}

# Check if two histograms are equal
hist_are_equal2() {
  # Array sizes differ?
  local size=${#_h1[*]}
  [ $size != ${#SUM[*]} ] && return 1

  # parsing input one index at a time
  for ((i=0; i < $size; i++))
  do
    [ ${_h1[i]} != ${SUM[i]} ] && return 1
  done

  return 0
}

# Filter out all words of incorrect length
# Use sum_word4 which generates a histogram of character frequency
alg6() {
  sum_word4 $word
  _h1=${SUM[*]}
  grep_string="^`echo -n $word | tr 'a-zA-Z' '.'`\$"

  grep "$grep_string" "$dict" | \
  while read line
  do
    sum_word4 $line
    if hist_are_equal
    then
      echo $line
    fi
  done
}

# Filter out all words of incorrect length
# Use sum_word4 which generates a histogram of character frequency
alg7() {
  sum_word4 $word
  _h1=${SUM[*]}
  grep_string="^`echo -n $word | tr 'a-zA-Z' '.'`\$"

  grep "$grep_string" "$dict" | \
  while read line
  do
    sum_word4 $line
    if hist_are_equal2
    then
      echo $line
    fi
  done
}

run_test() {
  echo alg$1
  eval time alg$1
}

#run_test 1
#run_test 2
#run_test 3
run_test 4
#run_test 5
run_test 6
#run_test 7

Answer 3

#!/usr/bin/perl
$myword=join("", sort split (//, $ARGV[0]));
shift;
while (<>) {
    chomp;
    print "$_\n" if (join("", sort split (//)) eq $myword);
}

像这样使用它： bla.pl < /usr/dict/words searchword

Answer 4

您想要查找仅包含给定字符集的单词。正则表达式是：

'^[letters_you_care_about]*$'

所以，你可以这样做：

grep "^[$W]*$" /usr/dict/words

'^'匹配行的开头; '$'表示行尾。这意味着我们必须具有完全匹配，而不仅仅是部分匹配（例如“表格”）。

'['和']'用于定义输入文件的一个字符空间中允许的一组可能字符。我们使用它来查找/ usr / dict / word中仅包含$ W中字符的单词。

'*'重复前一个字符（'[...]'规则），该字符用于查找任意长度的单词，其中所有字符均为$ W.

Answer 5

所以我们有以下内容：

n =输入字的长度
L =字典文件中的行

如果n往往很小而L往往很大，那么我们最好找到输入词的所有排列并寻找那些，而不是对字典文件的所有L行做某事（比如排序）？（实际上，因为找到一个单词的所有排列都是O（n！），并且我们必须为每个单词运行整个字典文件一次，也许不是，但我仍然编写了代码。）

这是Perl - 我知道你想要命令行操作，但我没有办法在shell脚本中做到这一点，而不是超级hacky：

sub dedupe {
    my (@list) = @_;
    my (@new_list, %seen_entries, $entry);

    foreach $entry (@list) {
        if (!(defined($seen_entries{$entry}))) {
            push(@new_list, $entry);
            $seen_entries{$entry} = 1;
        }
    }

    return @new_list;
}

sub find_all_permutations {
    my ($word) = @_;
    my (@permutations, $subword, $letter, $rest_of_word, $i);

    if (length($word) == 1) {
        push(@permutations, $word);
    } else {   
        for ($i=0; $i<length($word); $i++) {
            $letter = substr($word, $i, 1);
            $rest_of_word = substr($word, 0, $i) . substr($word, $i + 1);
            foreach $subword (find_all_permutations($rest_of_word)) {
                push(@permutations, $letter . $subword);
            }            
        }
    }

    return @permutations;
}

$words_file = '/usr/share/dict/words';
$word = 'table';

@all_permutations = dedupe(find_all_permutations($word));
foreach $permutation (@all_permutations) {
    if (`grep -c -m 1 ^$permutation\$ $words_file` == 1) {
        print $permutation . "\n";
    }
}

Answer 6

此实用程序可能会让您感兴趣：

an -w "tab" -m 3

...仅提供bat和tab。

原作者似乎不再存在，但您可以在http://packages.qa.debian.org/a/an.html找到信息（即使您不想自己使用它，也可能值得一看。）

在UNIX中查找包含字符的所有单词

6 个答案: