in reply to regular expression paranthesis remover

If you're unconcerned with their balancing except for the outer ones, 

$_ = '111(22(33)44)55';
my $re = qr/
    (.*?)  # minimal grab up to the first...
    \(     # literal left paren
    .*     # greedy skip everything up through...
    \)     # the last literal right paren
    (.*)$  # then grab everything remaining
    /x;
s/$re/$1$2/;
print;

[download]
If they do balance, it works, too. What do you want to happen if they don't balance?

Update: Typo repaired, thanks, Hofmator.

After Compline,
Zaxo

Replies are listed 'Best First'.
Re^2: regular expression paranthesis remover
by Hofmator (Curate) on Jun 25, 2004 at 09:19 UTC

    That's needlessly complicated, simply replace the stuff you don't want with nothing (as Anonymous Monk already said): s/\(.*\)//.

    The balancing caveat still applies, of course.

    Update: thospel is completely correct, my solution above (and Zaxo's) doesn't work for multiple parenthesis in the same string. To fix this (and actually make it recognise balanced parens): 

    $_ = "111(22(33)44)55";
1 while s/ \( [^()]* \) //gx;
if (/[()]/) {
  print "unbalanced!!";
} else {
  print;
}

    [download]
    BTW, I just noticed a small typo in Zaxo's code, it doesn't work as it stands. It should be my $re = qr/.../;.

    -- Hofmator

[reply]
[d/l]
[select]
      Fails on the perfectly balanced "12(34)56(78)9"

      Assuming it's about removing balanced parenthesis, I'd go for: 

      my $balance;
# Notice declaring $balance beforehand is important, 
# otherwise you pick up the global inside (??{})
$balance=qr/(?:[^()]|\((??{$balance})\))*/;
$string =~ s/\($balance\)//g;

      [download]
[reply]
[d/l]

In Section Seekers of Perl Wisdom