In order to compare plural values you have to normalize them - this means ordering them for comparison.

You have a minor error in your code - you're adding array counts, not elements, to the hash.

Start out by making an array of arrays: 
my @array = (
    [ sort @value1 ],
    [ sort @value2 ],
    # ... and so forth
);

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So that you have an array of normalized arrays, and then sort the large one and filter out duplicates like so: 
my $prev = [];
@array = grep {
    my $i = 0;
    my $ret = 0;
    $ret = 1 if $#$prev != $#$_;
    while ($ret && $i <= $#$prev and $i <= $#$_){
        $ret = 1 if $prev->[$i] ne $_->[$i];
    }
    $ret;
} sort {
    my $i = 0;
    my $ret = 0;
    loop: while (1) {
        if ($i <= $#{$a} and $i <= $#{$b}){ # test if we still have so
+mething to compute
            $ret = $a->[$i] cmp $b->[$i]; # compute the difference of 
+this element
            last loop if $ret; # if there's a difference return
            $i++;
        } else { # if the arrays are equal to this point in terms of e
+lements, but not in length, we decide on the longer one
            $ret = $#{$a} <=> $#{$b}; # determine by length
            last loop;
        }
    }
    $ret;
} @array;

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This will only return array unique arrays. Then you simply assign to your hash: 
$myhash{$key1} = $array[0];
$myhash{$key2} = $array[1];

[download]


-nuffin
zz zZ Z Z #!perl

In reply to Re: How to compare hash value by nothingmuch
in thread How to compare hash value by Anonymous Monk

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