On Feb 1, Robin Sheat said:
>On Fri, Jan 30, 2004 at 10:33:27AM -0500, Jeff 'japhy' Pinyan wrote:
>> sub array_diff {
>> my ($first, $second) = @_;
>> my %diff = map { "@$_" => 1 } @$second;
>> return [ map !$diff{"@$_"}, @$first ];
>> }
>amended to:
>return [ grep !$diff{"@$_"
On Fri, Jan 30, 2004 at 10:33:27AM -0500, Jeff 'japhy' Pinyan wrote:
> sub array_diff {
> my ($first, $second) = @_;
> my %diff = map { "@$_" => 1 } @$second;
> return [ map !$diff{"@$_"}, @$first ];
> }
amended to:
return [ grep !$diff{"@$_"}, @$first ];
(for the benefit of any
On Fri, Jan 30, 2004 at 10:33:27AM -0500, Jeff 'japhy' Pinyan wrote:
> 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].
Hmm, right. Not so good. I had thought of that, and thought I tested it
in that si
Robin Sheat wrote:
> Hey there, what is a nice way of doing what this looks like it should
> do:
>
> @a=([1,2,3],[5,5,5],[9,8,7]);
> @b=([5,5,5],[1,2,3]);
> @[EMAIL PROTECTED]@b;
>
> and have @c == ([1,2,3]);
>
> Is there a good way of doing this? (I've tried the obvious things on
> the command
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 @
On Fri, Jan 30, 2004 at 07:48:35AM -0700, Wiggins d Anconia wrote:
> Going to need more info about what you think this looks like it should
> do, because I (and maybe others here) lack the math skills to get your
> answer, funny, I got [9,8,7].
Thats what I get for not profraedinng, the result shou
>
> Hey there, what is a nice way of doing what this looks like it should
> do:
>
Going to need more info about what you think this looks like it should
do, because I (and maybe others here) lack the math skills to get your
answer, funny, I got [9,8,7].
> @a=([1,2,3],[5,5,5],[9,8,7]);
> @b=([5
