Re: Dot product full-sparse

[Marco Caliari]

On Thu, 12 Aug 2010, Jaroslav Hajek wrote:

> On Wed, Aug 11, 2010 at 8:20 AM, Marco Caliari <address@hidden> wrote:
> > Dear maintainers,
> >
> > if I run
> >
> > A = toeplitz ([-2, 1, 1, 1, zeros(1, 856)]);
> > A = reshape (A, 18490, 40);
> > SA = sparse (A);
> > B = rand (size (A));
> > tic, A .* B;,toc
> > tic, SA .* B;, toc
> > tic, SA .* sparse (B);, toc
> >
> > I get
> >
> > Elapsed time is 0.00745105 seconds.
> > Elapsed time is 0.229055 seconds.
> > Elapsed time is 0.0203849 seconds.
> >
> > (3.3.51 with Suitesparse 3.5.0, similar results with 3.2.4 with Suitesparse
> > 3.4.0). Why SA .* sparse (B) is so faster than SA .* B? What is done by
> > default when a full-sparse dot product is required?
> >
> > Best regards,
> >
> > Marco
>
> Apparently this operation is just poorly optimized.
>
> Before I start doing anything here, I'd like to know what the Matlab
> behavior is w.r.t. NaNs. Namely, if
> A = sparse ([2,3,1,2], [1,1,3,3], [0.1, 0.2, 0.3, 0.4]);
> B = ones (3); B(1,1) = NaN;
> C = sparse ([1,2,3,1,2], [1,1,1,3,3], [NaN, 0.1, 0.2, 0.3, 0.4]);
>
> what is the output of A.*B, B.*A, A ./ B, A .*C?
> Of course, newer version is better.

Here are the results with Matlab 7.6.0 (R2008a)

> > A.*B
ans =
   (1,1)          NaN
   (2,1)       0.1000
   (3,1)       0.2000
   (1,3)       0.3000
   (2,3)       0.4000
> >  B.*A
ans =
   (1,1)          NaN
   (2,1)       0.1000
   (3,1)       0.2000
   (1,3)       0.3000
   (2,3)       0.4000
> > A ./ B
ans =
   (1,1)          NaN
   (2,1)       0.1000
   (3,1)       0.2000
   (1,3)       0.3000
   (2,3)       0.4000
> > A .* C
ans =
   (1,1)          NaN
   (2,1)       0.0100
   (3,1)       0.0400
   (1,3)       0.0900
   (2,3)       0.1600

Marco