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how can i decrease the running time of for loop

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Dear All, I wrote this code to calculate accuracy of my work
but this code take about 2days to execution when the input is 5000 x 5000 (binary matrix),
so I want to minimize running time of my code.
2-result = zeros(u*2,2);
3-ri = 1;
4-for ii=1:u
5-for jj = ii+1:u
6-result(ri,:) = [ii jj];
7-ri = ri+1;
10-isRowToRemove = ismember(result,X,'rows');%test result matrix (5000 x 2) is a member in X (data matrix 4000 x 2) or not.
11-result(isRowToRemove,:) = [];
13-linindices = sub2ind(size(s), cc(:, 1), cc(:, 2))';% s is matrix(5000 x 5000)
14-b = s(linindices);%calculate the similarity of nonexistence links
16-linindices = sub2ind(size(s), B(:, 1), B(:, 2))';
17-A = s(linindices);%similarity of test links
18-ndash=sum(arrayfun(@(x) sum(x > b), A));%compare similarity of test and nonexistence links.
19-nddash=sum(arrayfun(@(x) sum(x == b), A));
20-nn=sum(arrayfun(@(x) sum(x < b), A));
21-auc=(ndash + 0.5 * nddash)/(ndash+nddash+nn);
suppose i have
X=[1 2
3 4
5 6]
represent the links between nodes 1,2,3,4 and 6.
lines from 1 to 12 calculate the remaining links of a complete network as
results=[1 3
1 4
1 5
1 6
2 3
2 4
2 5
2 6
3 5
3 6
4 5
4 6]
then in another code i calculate s (similarity that is about 5000 x 5000) then lines from 13 to 22 compare similarity(s) of b links (portion of X) and result links. this code take very long time about 48 hours without execution when X is about 5000 x 5000 matrix thus i want to minimize execution time
  1 Comment
Guillaume on 4 Apr 2018
Rather than leaving it up to us to decipher your code and understand what it is doing, why don't you explain what it's meant to do, in details?

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Accepted Answer

Guillaume on 4 Apr 2018
Without an explanation of what your code is doing it's near impossible to improve it. Particularly as you haven't explained how X, s and test are related. Saying that:
result = nchoosek(1:u, 2);
should be faster than your double for loop. It's certainly a lot shorter. Also,
ndash = sum(sum(A > b.'));
nddash = sum(sum(A == b.'));
nn = sum(sum(A < b.')); %or nn = numel(A) * numel(b) - ndash - nddash;
should be faster.
I don't understand the point of
cc = result;
B = test;
Why rename the variables? Why can't you use result and test in the rest of the code.
is pointless, since auc is guaranteed to be scalar.
muhammad ismat
muhammad ismat on 4 Apr 2018
You have already mentioned that in order to executed the basic code, you must have an s matrix and it calculated by another code and called in the basic code. I did not expect that the error come from the code of s so I did not write in the first time. It will be calculated from the following code
for i = 1:n %max(max(X))
for j = 1:i % <-- Note the 1:i instead of 1:n
s(i,j) = (DL(i,IDX(j))*DL(j,IDX(i))) / norm(DL(i,IDX(j))) * norm(DL(j,IDX(i)));
s(j,i) = s(i,j);
where the input for this code is matrix (5000 x 5000) i.e i= 1:5000 and also j, and each time calculate s(i,j) from mentioned equation that depend on DL (vector 5000 x 2)

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