Here's a rough draft (working properly, however) of one way to do it. I'm using a couple of hashes to reduce each array to simply a count of each element. Then the lesser of the two counts becomes the number of times set three should contain a given element. 

use strict;
use warnings;
use List::Util qw/min/;

my @list_a = qw/test test1 test2 test2 test/;

my @list_b = qw/test1 test2 test2/;

my( %element_count_a,
    %element_count_b
);

%element_count_a = %{ count( \@list_a ) };
%element_count_b = %{ count( \@list_b ) };


foreach my $key ( keys %element_count_a ) {
    foreach (   1 .. 
                min( $element_count_a{$key}, 
                exists( $element_count_b{$key} ) 
                    ? $element_count_b{$key} 
                    : '0' 
                ) 
    ) {
        print $key, "\n";
    }
}


sub count {
    my( $list_aref ) = shift;
    my %element_count;
    foreach my $item ( @{ $list_aref } ) {
        $element_count{$item}++;
    }
    return \%element_count;
}

[download]

Updated to choose better variable names.


Dave

In reply to Re: Compare Lists (Non Unique Intersection) by davido
in thread Compare Lists (Non Unique Intersection) by rsiedl

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