in reply to Find substring based on words and not in charachters

Update: Added minor optimisation. Update2: Rolled the optimisation into the while loop.

Something like this?:

#! perl -slw
use strict;

my $seg1 = "The man who likes reading books and writing poems.";
my $seg2 = "The man who likes reading big books and poems.";

my $best = '';
while( length( $seg1 ) > length( $best ) ) {
    while( $seg1 =~ m[(?!\s)(?=(\b.+\b)(?!\s))]g ) {
        my $bit = $1;
        $best = $bit if $seg2 =~ m[\Q$bit] and length( $bit ) > length
+( $best );
    }
    $seg1 =~ s[(?:\s|^)\S+$][];
}
print $best;

__END__
[17:04:21.46] C:\test>junk39
The man who likes reading

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  • Comment on Re: Find substring based on words and not in charachters (Updated.)
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Re^2: Find substring based on words and not in charachters (Updated.)
by choroba (Cardinal) on Dec 02, 2014 at 17:16 UTC
    I'd compare word count instead of length.
    لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ
[reply]

      The OP certainly could go that way, but I see two problems with it:

      1. Deciding upon a definition for a "word".
      2. Are a few short words more meaningful than 1 less long one?

        Eg. "on the way to" -v- "the Riechstag Bureau"?

      With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
      Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
      "Science is about questioning the status quo. Questioning authority".
      In the absence of evidence, opinion is indistinguishable from prejudice.
[reply]

      OP states he is looking for longest string; he wants to avoid getting parts of words, such as the last 'b' the lcss algorithm finds.

      1 Peter 4:10
[reply]
Re^2: Find substring based on words and not in charachters (Updated.)
by Anonymous Monk on Dec 03, 2014 at 09:20 UTC

    Hi,

    Thanks a lot for your code! It works perfect :)

[reply]

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