my %matches;
while ($str =~ /([abcd]{8,13})/g) {
    $matches{$1}++;
}

use Data::Dumper;
print Dumper \%matches;
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Hi,

Thanks for replying. I don't think your code is what I am looking for. I'll try to explain again:

For the sake of illustration let's say the size of the substring I am looking for is 3. Ideally, what I would like in the end is a list for all possible permutations with repetitions and a count how often they occur. For this example, the list would look something like that:

aaa 0
aab 2
aac 1
aad 0
aba 2
abb 0
abc 1
abd 0
etc.
How can this be achieved?
Thanks
Hadassa

Ideally, what I would like in the end is a list for all possible permutations with repetitions and a count how often they occur.

How often they occur... where? in which set/string/whatever?

If you mean in the list of possible permutations, why are some of them 0?

Perl 6 - links to (nearly) everything that is Perl 6.

Those four characters wouldn't happen to be a, g, t, c now would they? If so, you should head on over to the BioPerl site and check out some of the HOWTOs

Yes, indeed, I am dealing with DNA sequence. However, I could not find a suitable tool in BioPerl (I might have missed it, since this seems a pretty straight forward problem I have). Thanks for pointing this out.

An additional complexity to the problem is that I have some characters, say "x", in the string which are not to be counted, i.e.

aaaabccbaxxxcbddadcc...

So if n=3 as in the example any substring of size 3 that includes an x needs to be ignored and not counted.

Presumably, you don't want to count substrings that would be formed by the removal of the Xs?

Also, you mentioned earlier having all the permutations, including those with zero occurances, in the results hash. The problem with that is that by the time you get to n=13, you'd have 67 million entries in the hash, most of which would be 0.

Anyway, this might be of some use:

$s = 'aaaabccbaxxxcbddadcc';;

++$h{ $1 } while $s =~ m[(?=([abcd]{3})).]g;;

pp \%h;;
{
  aaa => 2,
  aab => 1,
  abc => 1,
  adc => 1,
  bcc => 1,
  bdd => 1,
  cba => 1,
  cbd => 1,
  ccb => 1,
  dad => 1,
  dcc => 1,
  dda => 1,
}
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---

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.

RIP an inspiration; A true Folk's Guy

Yes, forgot to mention that of course I will not store the one which have count zero.

Your answer was perfect! I guess I will have to catch up on some regular expressions!


Great!
Thanks a lot!

Very interesting solution. May I ask why you put two ;; at the end of each line and what pp in pp \%h" does ?

next if $substr =~ /x/;
[download]

Perl 6 - links to (nearly) everything that is Perl 6.

my $L = 4;
my @chars = qw[a b c d];
$_ = "aaaabdbbdaccdacdaaabdcd";

my %seen = map {$_, 0} glob do {local $" = ","; "{@chars}" x $L};
{local $" = ""; use re 'eval'; /([@chars]{$L})(?{$seen{$1}++})(*FAIL)/
+;}

while (my ($n, $v) = each %seen) {
    say "$n -> $v";
}
[download]

In my previous question in this thread I asked to count the number of occurrences of all permutations with repetitions. What if I only wanted to not have the count but a binary answer (i.e. 0 or 1) if a specific permutation is in my string? There must be a way to just change the regexp in a minor way.

$n=4;
$str="abababbcad";
%count;
while($str =~ m[(?=([abcd]{$n})).]g){
                   ++$count{ $1};     
               }
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$count{$1}=1
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$str1="abad";
$str2="aaab";
$n=2;
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