in reply to Re^2: Find common prefix from a list of strings
in thread Find common prefix from a list of strings

That approach is quite nice. Here's a unsightly twist on it: 
#!/usr/bin/perl -wl
use strict;
use warnings;
sub Prefix {
   my $p = shift;
   ($p)=$_=~join'','\A',map("(\Q$_",split//,$p),join'?',(')')x length(
+$p)or return'' for @_;
   $p
}

while (<DATA>) {
   print "$_=> ", Prefix(split ' ',$_);
}
__DATA__
model4run1 model2run1 model4run2 model1run1
model4run1 model2run1 model4run2 abbot
model4run1 model2run1 abbot model1run1
model4run1 model2run1 model4run2 monk
model4run1 model2run1 monk model1run1
testing terse testing time

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but I prefer the chop.

Replies are listed 'Best First'.
Re^4: Find common prefix from a list of strings
by slink (Initiate) on Feb 22, 2014 at 11:57 UTC

    I really liked the one-liner, just that it doesn't work - it fails for 000 720

    Here's what I came up with. Is doesn't look that efficient, but it has the advantage of using no regexen for the actual comparison (only for split). 
    #!/usr/bin/perl -wl
use strict;
use warnings;
sub Prefix {
    {
    my $e = scalar(@_);
    return '' if ($e == 0);
    return $_[0] if ($e == 1);
    }

    my $minlen;
    my @d = map { 
    my @split = split(//, $_);
    my $l = scalar(@split);
    if (defined($minlen)) {
        $minlen = $l if ($l < $minlen);
    } else {
        $minlen = $l;
    }

    \@split;
    } @_;

    my $p = '';
    my $i = 0;
    while($i < $minlen) {
    my $c = $d[0]->[$i];
    return $p unless ($c);
    for (my $j = 1; $j <= $#d; $j++) {
        return $p unless ($c eq $d[$j]->[$i]);
    }
    $p .= $c;
    $i++;
    }
        
   $p
}

while (<DATA>) {
   print "$_=> ", Prefix(split ' ',$_);
}
__DATA__
model4run1 model2run1 model4run2 model1run1
model4run1 model2run1 model4run2 abbot
model4run1 model2run1 abbot model1run1
model4run1 model2run1 model4run2 monk
model4run1 model2run1 monk model1run1
testing terse testing time
monk model4run1 model2run1 model1run1
000 720
a
aa
abc
a aa
a aa aaa
ab ba bb

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      You can also just add '^' as the second parameter to join in the previous answer.
      لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ
[reply]
[d/l]
[select]
      Thanks, fixed it (adding \A)
      --
      A math joke: r = | |csc(θ)|+|sec(θ)| |-| |csc(θ)|-|sec(θ)| |
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