Re: Print the oldest file in a directory.

Yeah, it dosen't make sense to me either. I had already written the little one liner when someone here mentioned using an ST instead. For the exact same reason you stated above. I am checking the mod time of the same file at a minimum of 2 times, maybe more if the next file makes it out of the <=> operation.

Please keep in mind, this is the first time I have ever used an ST. It is pretty much copied directly from http://www.5sigma.com/perl/schwtr.html.

Here is the benchmark code I used to test it with the results. The directory had 16,108 files in it. Let me know what you think.

#!/usr/bin/perl -w

use strict;
use Benchmark;

 timethese(100,
 { 'chad'   => \&chad,
   'swartz' => \&st,});

sub chad {
   sort{ (-M $b) <=> (-M $a) } glob("*");
}

sub st {
   map { $_->[0] }
   sort { $a->[1] <=> $b->[1] }
   map { [$_, -M] }
   glob('*');
}
[download]

Here are the results of the benchmark:

Benchmark: timing 100 iterations of chad, swartz...
      chad: 118 wallclock secs (56.91 usr + 47.25 sys = 104.16 CPU) @ 
+ 0.96/s (n=100)
    swartz: 1515 wallclock secs (142.33 usr + 1044.36 sys = 1186.69 CP
+U) @  0.08/s (n=100)
[download]

Comment on Re: Print the oldest file in a directory. Select or Download Code

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•Re: Re: Print the oldest file in a directory. by merlyn (Sage) on Oct 31, 2002 at 03:41 UTC
Doh! "glob *" is already sorted! Try again with a truly random order. -- Randal L. Schwartz, Perl hacker Be sure to read my standard disclaimer if this is a reply.	[reply]
Re: •Re: Re: Print the oldest file in a directory. by gnu@perl (Pilgrim) on Oct 31, 2002 at 16:02 UTC
Yup, DOH!. Technically I don't even need the <=>. I could just take the last output of glob. Still en expensive way of doing it. After reviewing the input here it would definately be better to perform an ST on an array populated using opendir/readdir. Thanks for your input everybody.	[reply]
Re: Re: Print the oldest file in a directory. by runrig (Abbot) on Oct 31, 2002 at 00:46 UTC
Methinks there is some kind of optimization going on. Perhaps perl knows that a sorted array in scalar (or void?) context is just the length of the array, and so the sort can be optimized away in the first case? Also, you really should take the glob out of the equation (it's an expensive operation and heavily skews the results of the benchmark). I get the ST being more than twice as fast with this (also you have '$a <=> $b' in one, and '$b <=> $a' in the other & BTW I have about 63 files in the directory): `use Benchmark; opendir(DIR, ".") or die "Acck: $!"; my @files = readdir DIR; closedir DIR; my $num_files = @files; print "$num_files\n"; timethese(-4, { 'chad' => \&chad, 'swartz' => \&st,}); sub chad { my @list = sort{ (-M $b) <=> (-M $a) } @files; } sub st { my @list = map { $_->[0] } sort { $b->[1] <=> $a->[1] } map { [$_, -M] } @files; }` [download]	[reply] [d/l]