On Jan 31, Robin Sheat said:
>> > @a=([1,2,3],[5,5,5],[9,8,7]);
>> > @b=([5,5,5],[1,2,3]);
>> > @[EMAIL PROTECTED]@b;
>#####
># subArray - takes two array references, and subtracts any instances
># of the arrays in the second one from the first one. Returns the new list.
>sub subArray {
> my @a = @{ shift() };
> my @b = @{ shift() };
> my @c;
> OUTER: foreach my $outer (@a) {
> INNER: foreach my $inner (@b) {
> next OUTER if ($outer == $inner);
> }
> push @c, $outer;
> }
> return @c;
>}
Well, that only works if $a[0] is [1,2,3] and $b[1] is $a[0] -- that is,
the EXACT SAME reference. It won't work if $b[1] is its own [1,2,3].
The simplest way to do this type of thing is to use a hash.
sub array_diff {
my ($first, $second) = @_;
my %diff = map { "@$_" => 1 } @$second;
return [ map !$diff{"@$_"}, @$first ];
}
This builds a hash of the elements to subtract, and then it goes through
the elements of the set being subtracted FROM, and returns only those
elements NOT in the subtract-me hash.
It might look convoluted, but it's efficient. It also prevents you from
doing the icky
for this (set a) {
for that (set b) {
compare this and that
}
}
--
Jeff "japhy" Pinyan [EMAIL PROTECTED] http://www.pobox.com/~japhy/
RPI Acacia brother #734 http://www.perlmonks.org/ http://www.cpan.org/
<stu> what does y/// stand for? <tenderpuss> why, yansliterate of course.
[ I'm looking for programming work. If you like my work, let me know. ]
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