in reply to Array definition error

What are you trying to do? 
$trainset[scalar(@trainset)] = \@inputs;
$trainset[scalar(@trainset)] = \@outputs;

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Is going to set the value to be \@outputs only (it will override).

So you should get 0 and 1 (the values of the outputs only) as your array elements.

Ok, nevermind, I'm thoroughly confused :)

I tested this and found the outputs to be the same except that the @trainset values are treated as strings. 

$VAR1 = [
          [
            '0',
            '0'
          ],
          [
            '0'
          ],
          [
            '0',
            '1'
          ],
          [
            '1'
          ]
        ];
$VAR1 = [
          [
            0,
            0
          ],
          [
            0
          ],
          [
            0,
            1
          ],
          [
            1
          ]
        ];

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Re^2: Array definition error
by gri6507 (Deacon) on Apr 15, 2005 at 15:25 UTC
    Actually that's not true. The line $trainset[scalar(@trainset)] = \@inputs; will increase the side of the @trainset array so that when $trainset[scalar(@trainset)] = \@outputs; is called, scalar(@trainset) will be one larger and therefore not overwrite the original data. This is akin to a push.
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[d/l]
[select]

      You statment is true. However, I'd like to ask why you are not simply using push? Does this method offer some benifit that I do not see? If so, please enlighten me.

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      —His Holiness, The Dalai Lama

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