Replace specific capture group instead of entire regex in Perl

Question

I've got a regular expression with capture groups that matches what I want in a broader context. I then take capture group $1 and use it for my needs. That's easy.

But how to use capture groups with s/// when I just want to replace the content of $1, not the entire regex, with my replacement?

For instance, if I do:

$str =~ s/prefix (something) suffix/42/

prefix and suffix are removed. Instead, I would like something to be replaced by 42, while keeping prefix and suffix intact.

score 20 · Answer 1 · answered Aug 26 '12 at 15:07

As I understand, you can use look-ahead or look-behind that don't consume characters. Or save data in groups and only remove what you are looking for. Examples:

With look-ahead:

s/your_text(?=ahead_text)//;

Grouping data:

s/(your_text)(ahead_text)/$2/;

score 3 · Accepted Answer · edited Feb 06 '14 at 14:28

3

If you only need to replace one capture then using @LAST_MATCH_START and @LAST_MATCH_END (with use English; see perldoc perlvar) together with substr might be a viable choice:

use English qw(-no_match_vars);
$your_string =~ m/aaa (bbb) ccc/;
substr $your_string, $LAST_MATCH_START[1], $LAST_MATCH_END[1] - $LAST_MATCH_START[1], "new content";
# replaces "bbb" with "new content"

edited Feb 06 '14 at 14:28

Flimm

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answered Aug 26 '12 at 15:09

Moritz Bunkus

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Would be anonymous downvoter please explain his vote? Thanks. – Moritz Bunkus Aug 26 '12 at 15:51

score 1 · Answer 3 · answered Apr 19 '13 at 17:39

This is an old question but I found the below easier for replacing lines that start with >something to >something_else. Good for changing the headers for fasta sequences

  while ($filelines=~ />(.*)\s/g){
        unless ($1 =~ /else/i){
                $filelines =~ s/($1)/$1\_else/;
        }

  }

score 1 · Answer 4 · answered Jun 27 '21 at 14:16

Use lookaround assertions. Quoting the documentation:

Lookaround assertions are zero-width patterns which match a specific pattern without including it in $&. Positive assertions match when their subpattern matches, negative assertions match when their subpattern fails. Lookbehind matches text up to the current match position, lookahead matches text following the current match position.

If the beginning of the string has a fixed length, you can thus do:

s/(?<=prefix)(your capture)(?=suffix)/$1/

However, ?<= does not work for variable length patterns (starting from Perl 5.30, it accepts variable length patterns whose length is smaller than 255 characters, which enables the use of |, but still prevents the use of *). The work-around is to use \K instead of (?<=):

s/.*prefix\K(your capture)(?=suffix)/$1/

score 0 · Answer 5 · answered May 12 '21 at 10:02

I use something like this:

s/(?<=prefix)(group)(?=suffix)/$1 =~ s|text|rep|gr/e;

Example:

In the following text I want to normalize the whitespace but only after ::=:

some    text     := a   b        c d   e   ;

Which can be achieved with:

s/(?<=::=)(.*)/$1 =~ s|\s+| |gr/e

Results with:

some    text     := a b c d e ;

Explanation:

(?<=::=): Look-behind assertion to match ::=

(.*): Everything after ::=

$1 =~ s|\s+| |gr: With the captured group normalize whitespace. Note the r modifier which makes sure not to attempt to modify $1 which is read-only. Use a different sub delimiter (|) to not terminate the replacement expression.

/e: Treat the replacement text as a perl expression.

Replace specific capture group instead of entire regex in Perl

5 Answers5