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I'd probably start by setting one array as the resulting intersection list, and comparing each array to that one.
my @arrays = ([...], [...], [...]); my %tmp_int; # assume intersection is entirety of first array # and set each "frequency" to the number of arrays @tmp_int{@{ $arrays[0] }) = (@arrays+0) x @{ $arrays[0] }; # then check the other arrays against it for my $set (@arrays[1..$#arrays]) { my %seen; $tmp_int{$_}-- for grep { !$seen{$_} } @$set; } # the ACTUAL intersection is all the # keys in the hash whose value is 1 my @intersection = grep { $tmp_int{$_} == 1 } keys %tmp_int;

Jeff japhy Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and perl hacker
How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart

In reply to Re: Getting the intersection of n collections. by japhy
in thread Getting the intersection of n collections. by jbrugger

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