Better way to use circshift

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Brent
Brent am 1 Nov. 2012
I am trying to create array B from array A (of course in the real application the numbers are not sequential and the arrays are much biggger i.e. 5000x5000)
B =
5 0 1 2 3 4
4 5 0 1 2 3
3 4 5 0 1 2
2 3 4 5 0 1
1 2 3 4 5 0
A =
0 1 2 3 4 5
0 1 2 3 4 5
0 1 2 3 4 5
0 1 2 3 4 5
0 1 2 3 4 5
Here is my solution:
for i = 1:5 B(i,:) = circshift(A(i,:),[0,i]); end
Is there a better (i.e. faster) way to do this? On a more general note, I get the fealing that using a for loop to act on a array is always a bad idea.
  2 Kommentare
Image Analyst
Image Analyst am 1 Nov. 2012
Wrong. Using for loops is not always a bad idea.
Brent
Brent am 2 Nov. 2012
Thanks for the answers. I didn't realize I would spark a debate. Unfortunatly, my array is not sparse. However, thaks to your comments, I realized that B is just a Toeplitz matrix based on the rows in A (duh ... can't believe I didn't see that). Here is the final code that I came up with:
r = rand(5000,1);
>> A = ones(size(r,1)-1,1)*r';
>> tic;B = toeplitz(fliplr(r'),circshift(r',[0,1]));toc
Elapsed time is 0.474317 seconds.
Not as good as Matt J., but it will do for my purpose. Thanks

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Antworten (2)

Matt J
Matt J am 1 Nov. 2012
Bearbeitet: Matt J am 1 Nov. 2012
If the matrix is going to be sparse, this should be pretty fast
B=interpMatrix([5 4 3 2 1],1,6,1,'circ');
B(end,:)=[];

Sean de Wolski
Sean de Wolski am 1 Nov. 2012
v = 0:5;
B = gallery('circul',v);
B = B(2:end,:);
A = repmat(v,[numel(v)-1,1]);
  3 Kommentare
Sean de Wolski
Sean de Wolski am 1 Nov. 2012
tic,B = gallery('circul',1:5000);toc
Elapsed time is 0.603616 seconds.
Matt J
Matt J am 1 Nov. 2012
It seems to be doing better in terms of memory management, compared to past MATLAB versions, but it's still slow. Take the example from my link. In R2012a, I get these timings:
r=[1 2 3 4 sparse(zeros(1,10000))];
tic; A=gallery('circul',r); toc
Elapsed time is 4.887724 seconds.
tic; A=interpMatrix(r,1,length(r),1,'circ').'; toc
Elapsed time is 0.005704 seconds.

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