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Re: vectorization
From: |
Francesco Potorti` |
Subject: |
Re: vectorization |
Date: |
Fri, 06 Dec 2002 10:50:35 +0100 |
>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
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-------------------------------------------------------------
- vectorization, David Pruitt, 2002/12/03
- Re: vectorization,
Francesco Potorti` <=