Re: Speedup/Refaktoring loop?

[Jaroslav Hajek]

On Tue, Aug 19, 2008 at 11:53 AM, Andreas Romeyke <address@hidden> wrote:
> Hi,
>
> There are any tricks to speedup the following piece of code?
>
>    for i=1:n
>        a=transpose(U) * X;
>        tmpU = transpose(U)*U;
>        for j=1:columns
>            b=tmpU*V;
>            V(i,j)=V(i,j)* a(i,j)/b(i,j);
>        end
>    end
>
> U and V are rectangular matrices, X is a squared matrix.
>
> The variant to replace line "V(i,j)=V(i,j)* a(i,j)/b(i,j);" with
>
> "V = V .* a ./ b" is much slower.
>
> I used Octave 2.9.9 to time the code.
>
> Any hints?
>
> Bye Andreas
>
>

If you really want to do what this code does, then I don't see any
speedup except hoisting "a" and "tmpU" out of the outer loop. But
chances are that you instead want something like this:

a = U.' * X;
b = U.' * U * V;
V = V .* a ./ b;

(*no* loops involved).

Note that this is not the same, since in your code the matrix b will
get a different value in each cycle of the inner loop.

>
> --
> Software Developer / Dipl. Inform. (FH)
> Max Planck Institute for Human Cognitive and Brain Sciences
> Department of Psychology
> Stephanstr. 1a, 04103 Leipzig, Germany
>
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-- 
RNDr. Jaroslav Hajek
computing expert
Aeronautical Research and Test Institute (VZLU)
Prague, Czech Republic
url: www.highegg.matfyz.cz