Re: parsing question

Unfortunately your question isn't terribly clear, so I'm not entirely sure what you're looking for.
However, one thing I would suggest - if you want to match after the last occurance of something, it may be easier to apply a regex to a reversed string, e.g:

my $reverse = reverse $line;
$reverse =~ s| \w* ; \s* > |>|x;
$line = reverse $reverse;
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Re: Re: parsing question by Chady (Priest) on May 28, 2003 at 09:30 UTC
or maybe a greedy regex? `$line =~ s/^(.>) ;./\1;/;` [download] He who asks will be a fool for five minutes, but he who doesn't ask will remain a fool for life. Chady \| http://chady.net/	[reply] [d/l]
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 ) }, } ); [download] 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.	[reply] [d/l]


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