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？
解决方法
看起来你需要一个
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
