Re: Question about speeding a regexp count

Another possibility:

$count{$1}++ while $seq =~ /(?=(.{1}))/g;
$count{$1}++ while $seq =~ /(?=(.{2}))/g;
$count{$1}++ while $seq =~ /(?=(.{3}))/g;
[download]

Update: woops, added the (?=) I initially forgot.

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Re^2: Question about speeding a regexp count by Vynce (Friar) on Oct 13, 2005 at 19:56 UTC
This approach actually does not work, because the regex starts matching after the previous match. in other words, if the $seq is "foobar" it will find 'foo' and then 'bar' but not 'oob' or 'oba'.	[reply]
Re^3: Question about speeding a regexp count by ikegami (Patriarch) on Oct 13, 2005 at 20:01 UTC
oops, I knew that! Thanks, Fixed.	[reply]