in reply to Re: How can I count characters between two same substrings in a string where the substring is repeated more than 5 times?
in thread How can I count characters between two same substrings in a string where the substring is repeated more than 5 times?

Elegant, but deviates from OP's spec, in that the character-count before the first "FOO" (or "cat") and after the last are reported in the output.

At risk of being bitten again by my blindspots, I see no way to force split alone (or even with a LIMIT) to do the job by itself. OTOH, adding a simple regex to remove the chars before the first delimiter and after the last would work nicely.

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Re^3: How can I count characters between two same substrings in a string where the substring is repeated more than 5 times?
by JavaFan (Canon) on Dec 09, 2011 at 13:34 UTC
    I'd use split. One of the points of the OP's spec is "repeated more than 5 times". A lot easier to check with split, than the cram it all in a regexp.

    Untested: 

    my @chunks = split /PAT/, $str;
shift @chunks unless $str =~ /^PAT/;
pop @chunks unless $str =~ /PAT$/;
if (@chunks > 4) {
    say "Lengths: @{[map {length} @chunks]}";
}

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      Since split discards an empty trailing string (but not a leading one), one option would be to add a character to the end that can't be part of the delimiter. Then you can split and take everything except the first and last elements: 

      #!/usr/bin/perl
use Modern::Perl;
my $a = 'catonecattwocatthreecatfourcat';
$a .= ' ';
my @w = split /cat/, $a;       
say "$_ -> " . length($_) for @w[1..@w-2];

one -> 3
two -> 3
three -> 5
four -> 4

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      Aaron B.
      My Woefully Neglected Blog, where I occasionally mention Perl.

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Re^3: How can I count characters between two same substrings in a string where the substring is repeated more than 5 times?
by tobyink (Canon) on Dec 10, 2011 at 14:32 UTC
    Yes, I deviated deliberately for simplicity. If you want to ignore the leading and trailing sections, just shift the first element off the list you get back from split, then pop the last element off as well.
[reply]
      That's all well and good, tobyink, but coding your proposed solution increases the complexity far beyond anything readily identifiable as "simple."

      To avoid complexity, one must know in advance that the delimiter follows something else at the start of the string (or is followed by something at the end)... in which case, counting with a pencil, paper and strike-thrus may be just as effective...
      or

      If the topology of the string is unknown, then you must deal with the permutations...

      • Delimiter first in the string, and last
      • Delimiter NOT first in string, but is last
      • Delimiter first in string, but NOT last
      • Delimiter neither first nor last
      • Delimiter occurs an odd number of times in string where it is either first or last
      • ... (I'm sure I've missed some)
[reply]

        Actually, if the delimiter occurs at the start of the string, then split returns an empty string as the first portion. If passed a LIMIT of -1, it will also do the same at the end of the string. So if you really don't care about the lengths of the "abc" and "xyz" portions, it's simply:

        use 5.010;
my $string = "abcFOOdefghFOOiFOOjklmFOOnopqrFOOstuvFOOwxyz";
my @portions = split /FOO/, $string, -1;
shift @portions;
pop @portions;
say "Lengths are:";
say $_ for map { sprintf("%d ('%s')", length, $_) } @portions;

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        As I said, the reason I didn't include the code to ignore the first and last portion of the string in my initial example is because it seemed obvious how it could be done.

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