Re: Array definition error

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

--------------
"But what of all those sweet words you spoke in private?"
"Oh that's just what we call pillow talk, baby, that's all."

Comment on Re: Array definition error Select or Download Code

Replies are listed 'Best First'.
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.	[reply] [d/l] [select]
Re^3: Array definition error by JediWizard (Deacon) on Apr 15, 2005 at 15:49 UTC
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. A truely compassionate attitude towards other does not change, even if they behave negatively or hurt you —His Holiness, The Dalai Lama	[reply]


Think about Loose Coupling
	PerlMonks