My problem is of course more complicated, and is obviously not a homework.
I just wanted to provide a minimal working example. You can replace the
matrix C with a matrix containing any number, for what matters. Btw,
because numbers are extracted from a Gaussian distribution, the likelihood
that you draw a 0 is actually zero.
Apart from this, apologies for having posted an html version.
On 17 December 2015 at 00:36, Jeff Newmiller <jdnew...@dcn.davis.ca.us>
wrote:
> This calculation divides by values centered around zero. The only context
> that I can think of that would require such silliness is a homework
> problem, and this list has a no-homework policy. If not, then mentioning
> the theory you are applying might help someone point you at an existing
> function that achieves your goals while avoiding divide-by-zero errors.
>
> Since you also posted in HTML I gather that you have not read the Posting
> Guide mentioned below. Avoiding HTML on this list is to your benefit, since
> using it inevitably leads to others seeing a garbled version of what you
> sent. Please read the PG for more important guidance.
> --
> Sent from my phone. Please excuse my brevity.
>
> On December 16, 2015 4:18:56 PM PST, Matteo Richiardi <
> matteo.richia...@gmail.com> wrote:
>
>> I have to evolve each element of a matrix W
>>
>> W <- matrix(0,2,3)
>>
>> according to some function which uses the indices of the matrix [i,j] as
>> arguments:
>> w.fun = function(i,j) {
>> return A[i]*B[j]/(C[i,j])
>> }
>>
>> where
>> A<-c(100,100)
>> B<-c(200,200,200)
>> C <- matrix( rnorm(6,mean=0,sd=1), 2, 3)
>>
>> How can I do it, without recurring to a loop? Also, in my application I
>> need to pass the function another argument.
>>
>> Thanks a lot for your suggestions.
>> Matteo
>>
>> [[alternative HTML version deleted]]
>>
>> ------------------------------
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>> and provide commented, minimal,
>> self-contained, reproducible code.
>>
>>
[[alternative HTML version deleted]]
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