Re: Efficient Octave Code

[Bill Denney]

rockster8 wrote:
> I've got a nested for-loop and was wondering if there is an octave-equivalent 
> notation that will simplify the for-loop:
>
> for i=1:50
>    for j=1:50
>       for k=1:10
>          Fk(i,j) = F(i,j) .* (w(i,j) == k);
>       end
>    end
> end
>
> where the following are initialized as:
> Fk(1:50,1:50) = 0;
> F(1:50,1:50) = 0;
> w(1:50,1:50) = randomly generated numbers..
>   
Well, first off, since you're multiplying 0 (F) times a boolean, Fk will 
always be zero, so

Fk = zeros (size (F));

would work.  Assuming that F weren't 0, you could do something like this:

for k = 1:10
 mask = (w == k);
 Fk(mask) = F(mask);
endfor

Bill