Re: Re: parsing question

or maybe a greedy regex?

$line =~ s/^(.*>) ;.*/\1;/;
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Comment on Re: Re: parsing question Download Code

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Re: Re: Re: parsing question
by kilinrax (Deacon) on May 28, 2003 at 11:06 UTC

In a word, no.

Reversing the regex is much faster.
Have a look at these benchmarks:

#!/usr/bin/perl -w

use strict;
use Benchmark;

my $string = "<<HTML>;nbsp dont_strip_me</HTML>> <xyzfdgfghgf> ;strip_
+me";

sub reversed {
  my $reverse = reverse(shift);
  $reverse =~ s| \w* ; \s* > |>|x;
  return scalar reverse $reverse;
}

sub greedy {
  my $line = shift;
  $line =~ s|^ (.*>) \s* ; \w* |$1|x;
  return $line;
}

print "Reversed: ", reversed($string), "\n";
print "Greedy: ", greedy($string), "\n";

timethese( -10,{
                reversed => sub { reversed( $string ) },
                greedy => sub { greedy( $string ) },
               } );
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Output:

Reversed: <<HTML>;nbsp dont_strip_me</HTML>> <xyzfdgfghgf> Greedy: <<HTML>;nbsp dont_strip_me</HTML>> <xyzfdgfghgf> Benchmark: running greedy, reversed, each for at least 10 CPU seconds... greedy: 10 wallclock secs ( 9.98 usr + 0.02 sys = 10.00 CPU) @ 78480.80/s (n=784808) reversed: 11 wallclock secs (10.46 usr + 0.00 sys = 10.46 CPU) @ 167660.04/s (n=1753724)

As you can see, it's over twice the speed. On longer strings, the difference would be even greater.

Also, your regex is wrong. Read through perldoc:perlre (specifically, the section marked 'Warning on \1 vs $1') to discover why.

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