毫无疑问,这对您来说很简单...
An easy one no doubt for you...
我有一个像这样的文件名列表;
I have a list of filenames which are like so;
fw_d.log.1.gz through fw_d.log.300.gz当我在下面的代码块中使用该代码块时,它几乎按照我想要的方式对其进行了排序,但并不完全.
When i use this below code block, it almost sorts it the way i want, but not quite.
#!/usr/bin/perl -w my $basedir = "/var/log"; my @verdir = qw(fw_d); my $fulldir; my $configs; my $combidir; foreach $combidir (@verdir) { $fulldir = "$basedir/$combidir"; opendir (DIR, $fulldir); my @files = grep { $_ ne '.' && $_ ne '..' && $_ ne 'CVS' readdir DIR; closedir (DIR); @files1 = sort {$a cmp $b}(@files); foreach my $configs (@files1) { print "Checking $configs\n"; system("less $basedir/$combidir/$configs | grep \'.* Group = , Username = .* autheauthenticated.\' >> output.log" ); } }这是代码段输出
Checking fw_d.log Checking fw_d.log.1.gz Checking fw_d.log.10.gz Checking fw_d.log.100.gz Checking fw_d.log.101.gz Checking fw_d.log.102.gz如您所见,它几乎按照我的希望进行排序....有人对我可以使用的阅读内容或代码段有任何建议吗?
As you can see, it almost sorts it how i was hoping.... Does anyone have any suggestions, on either reading, or a code snippet i can use?
预先感谢. 史蒂夫.
Thanks in advance. Steve.
推荐答案您可以使用 Schartzian变换:
my @sorted = map { $_->[0] } sort { $a->[1] <=> $b->[1] } map { [$_, $_=~/(\d+)/] } @files; print Dumper \@sorted;添加了用于比较Schwartzian-Transform和子例程的基准
Added benchmark for comparison between Schwartzian-Transform and subroutine
use Benchmark qw(:all); # build list of files my @files = map {'fw_d.log.'.int(rand()*1000).'.log' } 0 ..300; my $count = -3; my $r = cmpthese($count, { 'subname' => sub { sub expand { my $file=shift; $file=~s{(\d+)}{sprintf "%04d", $1}eg; return $file; } my @sorted = sort { expand($a) cmp expand($b) } @files; }, 'schwartzian' => sub { my @sorted = map { $_->[0] } sort { $a->[1] <=> $b->[1] } map { [$_, $_=~/(\d+)/] } @files; } });结果:
Rate subname schwartzian subname 21.2/s -- -92% schwartzian 279/s 1215% --Schwartzian转换对300个文件进行排序的效率提高了约13倍.
Schwartzian-transform is about 13 times more efficient for sorting 300 files.
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