c=5;
retint=0;
dist=log([10:-1:1]+retint);
for i=1:length(dist)
eta=exp(-c*abs(dist(i)-dist));
discrim(i)=1/sum(eta);
end
Does anyone know how to vectorise this for loop to make it more efficient?

댓글 수: 1

Because you are not concatenating anything, square brackets are not needed here:
dist=log((10:-1:1)+retint);

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 채택된 답변

David Hill
David Hill 2019년 12월 8일

0 개 추천

Although arrayfun is really a loop,
c=5;
retint=0;
dist=log([10:-1:1]+retint);
discrim=arrayfun(@(x)1/sum(exp(-c*abs(dist-x))),dist);

추가 답변 (1개)

Stephen23
Stephen23 2019년 12월 8일

0 개 추천

Real vectorized code (no loop or arrayfun):
eta = exp(-c*abs(bsxfun(@minus,dist,dist(:))));
discrim = 1./sum(eta,1)
Or for MATLAB versions >=R2016b:
eta = exp(-c*abs(dist-dist(:)));
discrim2 = 1./sum(eta,1)

카테고리

도움말 센터File Exchange에서 Loops and Conditional Statements에 대해 자세히 알아보기

질문:

2019년 12월 8일

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2019년 12월 8일

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