Another possible and less elegant way of traversing the column...

#!/usr/bin/perl

use strict;

my @results;
my @AoA = (
           [ "X", "Z", "W", "X", "Z", "Z" ],
           [ "Z", "Z", "X", "X", "X", "W" ],
           [ "X", "Z", "X", "W", "W", "W" ],
           [ "Z", "X", "W", "X", "Z", "Z" ],
           [ "Z", "X", "W", "X", "X", "W" ],
           [ "Z", "X", "X", "X", "Z", "W" ],
);

for my $i ( 0 .. $#AoA ) {
    my %count;
    for my $j ( 0 .. scalar( @{ $AoA[0] } ) - 1) {
        $count{ $AoA[$j]->[$i] }++;
    }

    for my $k ( reverse sort { $count{$a} <=> 
      $count{$b} } keys %count ) {
        print "$k\n";
        last;
    }
}

print join ", ", @results;
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