#!/usr/bin/perl -w

use strict;

my $string = "abacab";
print "$string\n";

my $pattern = "a";
my $better = "A";
my $n=1;
my $i=0;

$string =~ s/($pattern)/$i == $n ? $better : $1; i++/ge;

print "$string\n"
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Nuance

Black magic! I'm really impressed, I didn' know you can put that all into the substitution pattern. /me bows to Nuance ...

As for my solution, I was focused on the sample input given, I didn't think of the more general problem. But you're right about this.

As for your script, it works with a small alteration:

$string =~ s/($pattern)/$i++ == $n ? $better : $1/ge;
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If you increment $i in a seperate statement, the whole block evaluates to the value of $i und you get

abacab
0b1c2b
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Ok, now we're having fun :-)...

Andreas (waiting for the chinese food delivery service ...)

To make it even clearer, remember that the right side of the s///ge can be an arbitrary block. Here's the "Perl Bowling" style written for clarity:

{ my $count = 0;
  $string =~ s{($pattern)}{
    my $source = $1;
    $count++;
    if ($count == $target) {
      $source = $replacement; # replace!
    }
    $source;  # either original or replacement now
  }ge;
};
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And yes, you can actually write it like that, just as you see. If you're doing anything odd in the right hand side in a s///e, I recommend expanding it out like that.

-- Randal L. Schwartz, Perl hacker

Thanks, your solution works great! I wrote a sub to convert sed-ish s///n constructs to your format, making conversion from sed easier:

sub seval
{
    my ($sub) = shift;
    $vni++;
    $sub =~ s#s/([^/]+)/([^/]+)/(.+)#s/\1/\$i$vni++ == \3 ? \2 : \$&/e
+gs#;
    return $sub;
}
# Evals 's/foo/$i1++ == 1 ? bar : $&/egs'
eval(seval("s/foo/bar/2"));
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Thanks again.

I was going to give a similar solution, but then I realised that the previous (non-replaced) matches didn't necessarily have to be consecutive in the original string. Perhaps you you need to put optional separator space in the regex. Something like...

$string =~ s/^(($pattern.*){$n})$pattern/$1$better/;
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The .* is greedy, as far as I can see that will always replace ithe last match if you have more than the number you need.

my string = "foo_a_foo_b_foo_c_foo_d_foo_e_foo_f_foo"
my $pattern = "foo"
my $better = "bar"
my $n = 1;

$string =~ s/^(($pattern.*){$n})$pattern/$1$better/;
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would give:

"foo_a_foo_b_foo_c_foo_d_foo_e_foo_f_bar"
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not

"foo_a_bar_b_foo_c_foo_d_foo_e_foo_f_foo"
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Nuance

You're absolutely right, of course. Now I remember why I didn't post that solution in the first place :(

Update:

On thinking about it further, I realise that:

$string =~ s/^(($pattern.*?){$n})$pattern/$1$better/;
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will probably do the trick.

--
<http://www.dave.org.uk>

European Perl Conference - Sept 22/24 2000, ICA, London
<http://www.yapc.org/Europe/>