arrow,
I think you left out some important information. As BrowswerUK pointed out - are you looking for a match if the sequence doesn't match, but all the elements are present? Is a match acceptable if the same element is repeated in one array, but is only present once in the other? If you have a requirement to match duplicates than converting to a hash without a count is not going to work.

#!/usr/bin/perl -w
use strict;
my @array1 = (1,2,2,4);
my @array2 = (2,2,4,1,6);
my %hash1;
my %hash2;
my $not_ok;
foreach(@array1){
    $hash1{$_} +=1;
}
foreach(@array2){
    $hash2{$_} +=1;
}
while (my ($key,$hash1_value) = each %hash1) {
    if ($hash2{$key} && ($hash1_value == $hash2{$key})) {
        next;        
    }
    else {
        $not_ok = 1;
        last;
    }
}
if ($not_ok) {
    print "Array1 is not a subset of Array2\n";
}
else {
    print "We have a winner\n";
}
[download]

There is a penalty in efficiency for wanting this kind of accuracy - thanks to gmax for pointing this out and my solution using Data::Dumper was just plain wrong!
Cheers - L~R

Update: If sequence is important, then I would use the following logic:

Determine the smaller of the two arrays

Store the first element of the smaller array in a variable

Do a for loop on the larger array in the (0 .. $#array) form

Check each element for the first element of smaller array or next

Upon match, verify each subsequent element is a match

Match if you reach the last element of smaller array

No match if you don't

There is also a module that can preserve the order of a tied hash through some unknown magic you could look at, but I don't know much about it.

Comment on Re: Matching arrays Download Code