"?" is taken to be the start of a ?...? operator, but there's no other "?" before the end of the file to end the operator, thus the error.

You seem to think ?( EXPR ){/w,6}; is valid Perl, but nothing about it is even close to Perl.

Just guessing, but maybe something like this?

my $temp_seq = "fooXXXXXXbar";
my $tag_seq  = "foo";

$temp_seq =~ s/^($tag_seq)\w{6}(.*)/$1$2/;

print $temp_seq;   # "foobar"
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(removes the six characters following the search pattern "foo", which must occur at the beginning of $temp_seq)

See s/// under Regexp-Quote-Like-Operators.

Another way to achieve the same would be

if ($temp_seq =~ /^$tag_seq/) {
    substr($temp_seq, $+[0], 6) = '';
}
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where $+[0] is the position after the matched pattern.

Or:  s/^($tag_seq)\w{6}/$1/
Or (requires 5.10 due to use of \K):  s{ \A $tag_seq \K \w{6} }{}xms
Or (requires 5.10 due to use of \K):  s{ \A \Q$tag_seq\E \K \w{6} }{}xms
Or...

Thanks for the help! However, I would like to do a little bit more than just remove 6 characters following the pattern; I would like to remove the pattern as well as the 6 preceding/following characters.

The following did remove the 6 following characters but I would like to integrate the pattern into the regex so that it will also be removed.:


$temp_seq =~ s/^($tag_seq)\w{6}(.*)/$1$2/;
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Any suggestions?

I would like to remove the pattern as well as the 6 preceding/following characters.

Actually, that's even easier; see my other reply.

Could you clearify what it is you want?

is there a way to invert the pattern then apply it?

Forgive me for not being as clear as I thought I was.
What I have working now (as seen in the code in my last post) is a pattern that is removed (in this case it's replaced by an asterisk for testing purposes) in $temp_seq and I would like to have the preceding or following 6 characters removed as well.

For example:


$temp = 1234567890;
$pattern = 123;

$temp =~ s/$pattern/*/i;

print "temp = $temp";
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The output I would like to see is: temp = 0

Like wise, if the pattern removed was at the end of the string, I would like to remove the preceding 6 characters of the $temp.
As for the inversion of the pattern, consider the example mentioned earlier, where the pattern removed was found at the end of the string but it was inverted:

$temp = 4567890321;
$pattern = 123;
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Instead of creating a whole new variable to match the 321, is there a simpler way to reverse the $pattern variable explicitly in a regex?

please, let me know if I need to re-clarify anything and thanks again!!

The output I would like to see is: temp = 0

As whatever matches will be replaced, just include those 6 characters in the search pattern:

$temp =~ s/$pattern\w{6}//i;
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As for the inversion, see reverse.

I would probably write that as:

my $pattern = '...';
$temp =~ s/${pattern}.{6}//s || $temp =~ s/.{6}${pattern}//s;
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Not sure you need the /i, as your example uses numbers.

Instead of creating a whole new variable to match the 321, is there a simpler way to reverse the $pattern variable explicitly in a regex?

But what if the pattern is more complicated? And why do you consider creating a "whole new variable" such a big deal? I guess if your pattern is just a simple string without characters special to the regexp engine, you could write

$temp =~ s/${\scalar reverse $pattern}.{6}//s ||
$temp =~ s/.{6}${\scalar reverse $pattern}//s;
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but I wouldn't. The following is, IMO, much simpler:

my $rpattern = reverse $pattern;
$temp =~ s/$rpattern.{6}//s || $temp =~ s/.{6}$rpattern//s;
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Just playing around with Perl6 some

#!/home/ejh/rakudo/perl6

my @temp = ("1234567",
            "7654321");

my $pat  = "123";
my $rpat = "321";


for @temp -> $test {
        print "Testing '$test' - ";
        my $t = $test;
        if $t ~~ s/$pat// or $t ~~ /$rpat/ {
                say "($t) matched";
        }
}
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Thank you for all of your help! I finally got it working! However, I have another question. I'm not very familiar with lookarounds but I think that they would increase the throughput of my program exponentially!

I would like to use a lookahead (and/or lookbehind, I don't really know the difference..) assertion after matching a pattern (from a list of patterns) at either the beginning or end of the string. The lookaround would search for an adjacent pattern that is also from the list (NOTE: order does not matter!). If there is no known adjacent pattern, then remove the 6 characters preceding or following the pattern (as we have done already); if there is a pattern, proceed with some other instruction (of which is not relevant). The following is pseudocode to help explain:

$pattern_A = "123";
$pattern_b = "456";

$string_1 = "1234567890";
$string_2 = "1236547890";

# Then apply regex/lookaround to search for adjacent pattern

if($adjacent_pattern == false){
     remove_pattern_and_six_following_characters;
     print("string_*");
}
else{
     do_something_else;
     print("string_*");
}
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