Optimise/replace this for loop used in calculation of mutual information
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So I previously posted on here with regards to some code given HERE. Thanks to the help of Jan Simons I have managed to eliminate one of the nested for loops to speed up the code pretty significantly. However, due to the dimensionality of the project and future projects I am working on I still need to optimise the code. I ran the profiler and the following block of code seems to be bottleneck:
Eps = zeros(nObs, 1);
Nn = zeros(nObs, 1);
nx1 = zeros(nObs, 1);
ny1 = zeros(nObs, 1);
nx2 = zeros(nObs, 1);
ny2 = zeros(nObs, 1);
for i = 1:nObs
dxSample = dx(i, :);
dxSample(i) = [];
dySample = dy(i, :);
dySample(i) = [];
dzSample = dz(i, :);
dzSample(i) = [];
[EpsSample, NnSample] = sort(dzSample, 'ascend');
Eps(i) = EpsSample(k);
Nn(i) = NnSample(k);
nx1(i) = sum(dxSample < Eps(i));
ny1(i) = sum(dySample < Eps(i));
nx2(i) = sum(dxSample <= Eps(i));
ny2(i) = sum(dySample <= Eps(i));
end
Note that the k is typically quite low: between 3-10, nObs is around 4364. Is there any way of speeding this up? Maybe even eliminating the for loop altogether if possible? Any help is very much appreciated!
5 个评论
Walter Roberson
2018-1-16
I have seen several reports that implicit expansion is often slower than bsxfun. The details are likely to vary with release.
采纳的回答
Greg
2018-1-16
编辑:Greg
2018-1-18
Taking some guesses:
eyeinds = eye(nObs,'logical');
dx2(eyeinds) = Inf;
dy2(eyeinds) = Inf;
dz2(eyeinds) = Inf;
% Edit to incorporate Jan's suggestions:
[EpsSample2, NnSample2] = mink(dz2,k,2);
Eps2 = EpsSample2(:,end);
Nn2 = NnSample2(:,end);
% [EpsSample2, NnSample2] = sort(dz2,2,'ascend');
% Eps2 = EpsSample2(:,k);
% Nn2 = NnSample2(:,k);
nx12 = sum(dx2 < Eps2,2);
ny12 = sum(dy2 < Eps2,2);
nx22 = sum(dx2 <= Eps2,2);
ny22 = sum(dy2 <= Eps2,2);
更多回答(1 个)
Jan
2018-1-16
Sorting the complete array is a waste of time if all you need is the k.th smallest element. Under Matlab 2017b see mink to solve this much faster.
A further idea: Omit setting the diagonal to Inf, but leave it at zero. Then the result of nx12 etc. is 1 to large.
[Eps2, Nn2] = mink(dz2, 2, k);
nx12 = sum(dx2 < Eps2, 2) - 1;
ny12 = sum(dy2 < Eps2, 2) - 1;
nx22 = sum(dx2 <= Eps2, 2) - 1;
ny22 = sum(dy2 <= Eps2, 2) - 1;
2 个评论
Greg
2018-1-16
编辑:Greg
2018-1-17
Leaving at zero only works if all data is guaranteed positive. And that throws off the min counting in dz. I didn't want to make that assumption.
Just noticed your version leaves Eps2 as a non-vector for k > 1. I've edited my answer to incorporate the awesome suggestion to use mink.
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