Can someone do this calculation without for loops ?
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a = [1 2 3; 4 5 6];
b = [ 1 2 3];
for n = 1: size(a,1)
for m = 1:size(a,2)
k(n,m,:)= b.*b*a(n,m)
end
end
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Accepted Answer
More Answers (2)
Joakim Magnusson
on 15 Jun 2016
Do you mean like this?
fun=@(a,b) b.*b*a
k = bsxfun(fun,a,b)
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Azzi Abdelmalek
on 15 Jun 2016
Edited: Azzi Abdelmalek
on 15 Jun 2016
a = [1 2 3; 4 5 6];
b = [ 1 2 3];
bb=reshape(b.*b,1,1,[])
out=bsxfun(@times,a,bb)
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