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Rearranging of matrix in order to avoid loop

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Thomas G
Thomas G am 24 Jul. 2017
Kommentiert: Jan am 25 Jul. 2017
In the following code, my goal is to avoid the for-loop (i=1:m). I have already tried several things, but all of them failed.
m = 3;
n = 10;
ii = [1:n;3:n+2;2:n+1]; % index vector with size (m,n)
jj = [3:n+2;2:n+1;1:n]; % index vector with size (m,n)
load E.mat %external file see attachment
res = zeros(n,n,m);
for i=1:m;
res(:,:,i) = E(ii(i,:),jj(i,:),1,1);
end
How can I avoid the loop? In the attachment you can find the reference output matrix res.mat which contains the matrix res generated by this code. A code without the loop should reproduce the result saved in res.mat.
  7 Kommentare
Stephen23
Stephen23 am 24 Jul. 2017
Bearbeitet: Stephen23 am 24 Jul. 2017
@Thomas G: I assumed that you are not using random numbers in your calculations, which is why I did not put that as an answer.
However, your question, as posed, is best answered by a simple rand call. Given that you use random numbers to fill that matrix it is impossible to write and test code and know if the matrix has been filled correctly: one random matrix is just as good as another for solving the question that you have asked. Put another way, your question is impossible to falsify: there is no way that we could write code that is wrong, as long as it generates some random numbers. We need to be able to check if code output is wrong.
What would be actually useful is to provide small input and output examples: matrices that actually show what you want to achieve, and that can be used to test code on. You can upload data files by clicking the paperclip button.
Thomas G
Thomas G am 25 Jul. 2017
Ok thank you for your hint. I modified the question and added an input and an output file for my loop example. Hope that makes it now precise.

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Akzeptierte Antwort

Stephen23
Stephen23 am 25 Jul. 2017
Bearbeitet: Stephen23 am 25 Jul. 2017
>> II = repmat(permute(ii,[2,3,1]),1,10,1);
>> JJ = repmat(permute(jj,[3,2,1]),10,1,1);
>> rex = E(sub2ind(size(E),II,JJ));
>> isequal(rex,res)
ans = 1
  5 Kommentare
Stephen23
Stephen23 am 25 Jul. 2017
@Thomas G: the only way to know for sure is to test it. Write both versions in functions and time them using timeit. You might like to read these as well:
Jan
Jan am 25 Jul. 2017
@Thomas G: As a rule of thumb vectorized code is faster than loops, if no huge intermediate index or data arrays are created. When the data process in a loop match into the processor cache, this is a big advantage, and the best vectorized method is lame, if virtual memory must be used, because the physical RAM is exhausted.
Matlab does not parallelize magically. But some commands are multi-threaded, e.g. sum, filter, min, etc. Above a certain size of inputs the work is distributed to the available cores. You can check this roughly in the task manager.

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Saeed Bello
Saeed Bello am 25 Jul. 2017
You can try to look up Matrix indexing and Vectorization in MATLAB. Hope this help.

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