Question

假设我有一个数组，@ theArr，它包含大约1,000个元素，如下所示：

01  '12 16 sj.1012804p1012831.93.gz'
02  '12 16 sj.1012832p1012859.94.gz'
03  '12 16 sj.1012860p1012887.95.gz'
04  '12 16 sj.1012888p1012915.96.gz'
05  '12 16 sj.1012916p1012943.97.gz'
06  '12 16 sj.875352p875407.01.gz'
07  '12 16 sj.875408p875435.02.gz'
08  '12 16 sj.875436p875535.03.gz'
09  '12 16 sj.875536p875575.04.gz'
10  '12 16 sj.875576p875603.05.gz'
11  '12 16 sj.875604p875631.06.gz'
12  '12 16 sj.875632p875659.07.gz'
13  '12 16 sj.875660p875687.08.gz'
14  '12 16 sj.875688p875715.09.gz'
15  '12 16 sj.875716p875743.10.gz'
...

如果我的第一组数字（在'sj。'和'p'之间）总是6位数，我就不会有问题。但是，当数字翻转为7位数时，默认排序将停止工作，因为较大的7位数字位于较小的6位数字之前。

有没有办法告诉Perl按每个数组元素中字符串内的那个数字排序？

Answer 1

您似乎需要Schwartzian Transform：

#!/usr/bin/perl

use strict;
use warnings;

my @a = <DATA>;

print 
    map  { $_->[1] }                #get the original value back
    sort { $a->[0] <=> $b->[0] }    #sort arrayrefs numerically on the sort value
    map  { /sj\.(.*?)p/; [$1, $_] } #build arrayref of the sort value and orig
    @a;

__DATA__
12 16 sj.1012804p1012831.93.gz
12 16 sj.1012832p1012859.94.gz
12 16 sj.1012860p1012887.95.gz
12 16 sj.1012888p1012915.96.gz
12 16 sj.1012916p1012943.97.gz
12 16 sj.875352p875407.01.gz
12 16 sj.875408p875435.02.gz
12 16 sj.875436p875535.03.gz
12 16 sj.875536p875575.04.gz
12 16 sj.875576p875603.05.gz
12 16 sj.875604p875631.06.gz
12 16 sj.875632p875659.07.gz
12 16 sj.875660p875687.08.gz
12 16 sj.875688p875715.09.gz
12 16 sj.875716p875743.10.gz

Answer 2

您可以使用正则表达式从您传递给排序函数的块内的每一行中提取数字：

@newArray = sort { my ($anum,$bnum); $a =~ /sj\.([0-9]+)p/; $anum = $1; $b =~ /sj\.(\d+)p/; $bnum = $1; $anum <=> $bnum } @theArr;

但是，查斯。 Owens的解决方案更好，因为它只对正则表达式匹配每个元素一次。

Answer 3

这是一个将他们升序排序的例子，假设你不太关心效率：

use strict;

my @theArr = split(/\n/, <<END_SAMPLE);
12 16 sj.1012804p1012831.93.gz
12 16 sj.1012832p1012859.94.gz
12 16 sj.1012860p1012887.95.gz
12 16 sj.1012888p1012915.96.gz
12 16 sj.1012916p1012943.97.gz
12 16 sj.875352p875407.01.gz
12 16 sj.875408p875435.02.gz
12 16 sj.875436p875535.03.gz
12 16 sj.875536p875575.04.gz
12 16 sj.875576p875603.05.gz
END_SAMPLE

my @sortedArr = sort compareBySJ @theArr;

print "Before:\n".join("\n", @theArr)."\n";
print "After:\n".join("\n", @sortedArr)."\n";

sub compareBySJ {
    # Capture the values to compare, against the expected format
    # NOTE: This could be inefficient for large, unsorted arrays
    #       since you'll be matching the same strings repeatedly
    my ($aVal) = $a =~ /^\d+\s+\d+\s+sj\.(\d+)p/
        or die "Couldn't match against value $a";
    my ($bVal) = $b =~ /^\d+\s+\d+\s+sj\.(\d+)p/
        or die "Couldn't match against value $a";

    # Return the numerical comparison of the values (ascending order)
    return $aVal <=> $bVal;
}

输出：

Before:
12 16 sj.1012804p1012831.93.gz
12 16 sj.1012832p1012859.94.gz
12 16 sj.1012860p1012887.95.gz
12 16 sj.1012888p1012915.96.gz
12 16 sj.1012916p1012943.97.gz
12 16 sj.875352p875407.01.gz
12 16 sj.875408p875435.02.gz
12 16 sj.875436p875535.03.gz
12 16 sj.875536p875575.04.gz
12 16 sj.875576p875603.05.gz
After:
12 16 sj.875352p875407.01.gz
12 16 sj.875408p875435.02.gz
12 16 sj.875436p875535.03.gz
12 16 sj.875536p875575.04.gz
12 16 sj.875576p875603.05.gz
12 16 sj.1012804p1012831.93.gz
12 16 sj.1012832p1012859.94.gz
12 16 sj.1012860p1012887.95.gz
12 16 sj.1012888p1012915.96.gz
12 16 sj.1012916p1012943.97.gz

Answer 4

是。 sort函数采用可选的比较函数，用于比较两个元素。它可以采用代码块或要调用的函数名称的形式。

链接文档中有一个与您要执行的操作类似的示例：

# inefficiently sort by descending numeric compare using
# the first integer after the first = sign, or the
# whole record case-insensitively otherwise

@new = sort {
($b =~ /=(\d+)/)[0] <=> ($a =~ /=(\d+)/)[0]
            ||
            uc($a)  cmp  uc($b)
} @old;

使用Perl，如何使用每个数组元素中的数字值对数组进行排序？

4 个答案: