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Re: vectorization
From: |
Paul Kienzle |
Subject: |
Re: vectorization |
Date: |
Fri, 6 Dec 2002 09:46:01 -0500 |
User-agent: |
Mutt/1.2.5.1i |
On Fri, Dec 06, 2002 at 10:50:35AM +0100, Francesco Potorti` wrote:
> >In some work I'm now doing, I continually find the need to determine the
> >set of indices of vector A which contain elements of vector B - elements of
> >B will occur in A no more than once. Here's an example of what my code
> >looks like:
> >
> ># contrived example
> ># vector A
> >i_num = [1:100];
> ># vector B
> >v_num = [ 1 3 5 12 ];
> >
> >lt = length(v_num);
> >
> >for i = 1:lt
> >
> > idx(i) = find( i_num == v_num(i) );
> >
> >endfor;
> >
> >In this example, idx would of course be 1,3,5,12. So it's sort of a
> >two-dimensional "find". I'll just bet I'm doing this the hard way. Is
> >there another way without a for-loop? For the cases I'm seriously
> >concerned with, the vectors contain 5,000 to 15,000 elements.
>
> I tried to do it in two dimensions, by building ad hoc matrices and
> comparing them. This way, I hoped to trade memory space for cpu cycles.
> Unfortunately my way is twice as slow as yours. Maybe on a different
> architecture this will change, or maybe someone will have a better idea
> looking at my example:
>
> octave2.1> v=[504030,1566234,181111,32383]; n=[0:1500000]';
> octave2.1> tic; for i=1:length(v); idx(i)=find(n==v(i)); endfor; sort(idx)',
> toc
> ans =
>
> 11112 32384 181112 504031
>
> ans = 3.5039
> octave2.1> tic; find(sum((ones(size(n))*v==n*ones(size(v)))')), toc
> ans =
>
> 32384 181112 504031
>
> ans = 7.3894
If n contains valid indices, you could also be able to do it with sparse:
n=[2:2:150000];
v=[4:16:100000];
tic; sparse(1,n,[1:length(n)])(v), toc
ans = 0.32846
but it doesn't get you that much beyond the more general "lookup" solution:
tic; lookup(n,v); toc
ans = 0.42895
Paul Kienzle
address@hidden
>
> --
> Francesco PotortÃ¬ (researcher) Voice: +39 050 315 3058 (op.2111)
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>
>
>
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