Is it possible to vectorize this ?

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Omar Ali Muhammed
Omar Ali Muhammed 2021-4-26
编辑： Matt J 2021-4-26
A matrix A(100x1000) a selective column defined in B(1x1000). The required processing is the mean of non zero elements of each column of A referenced by B. Without the use of loops.
Regards

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采纳的回答

Matt J
Matt J 2021-4-26
编辑：Matt J 2021-4-26
A(~A)=nan;
result=mean(A(:,B),1,'omitnan')
  2 个评论
Omar Ali Muhammed
Omar Ali Muhammed 2021-4-26
Dear, it is wonderful.
How can we 'omitnan' if we replace mean by trimmean?
Matt J
Matt J 2021-4-26
编辑：Matt J 2021-4-26
Dear, it is wonderful.
I'm very glad. Please Accept-click the answer to indicate so.
How can we 'omitnan' if we replace mean by trimmean?
What would you be excluding?Zeros again? If so,
A(~A)=inf;

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更多回答（1 个）

Jan
Jan 2021-4-26
AB = A(:, B);
result = sum(AB, 1) ./ sum(AB ~= 0, 1);
mean('omitnan') replaces the NaNs by zeros for the summation and calculates the number of non-NaNs by sum(~isnan(AB)). Therefore this code should be faster, because it avoids replacing zeros by NaNs and back to zeros again.
  1 个评论
Matt J
Matt J 2021-4-26
编辑：Matt J 2021-4-26
because it avoids replacing zeros by NaNs and back to zeros again.
+1. Although, I would argue, they probably should have been NaNs instead of zeros from the very beginning.

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