Can someone do this calculation without for loops ?
显示 更早的评论
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
采纳的回答
k = bsxfun(@times, a , reshape(b.^2,1,1,[]))
alt_k = bsxfun(@times, a , permute(b.^2,[3,1,2]))
更多回答（2 个）
Joakim Magnusson
2016-6-15
Do you mean like this?
fun=@(a,b) b.*b*a
k = bsxfun(fun,a,b)
Azzi Abdelmalek
2016-6-15
编辑：Azzi Abdelmalek
2016-6-15
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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