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Can someone do this calculation without for loops?

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Amelos
Amelos 2016-6-20
评论： Stephen23 2016-6-21
a = [1 2 3; 4 5 6];
b = [ 1 2 3 4];
c = [1 2 3 4 5];
for n = 1: size(a,1)
for m = 1:size(a,2)
for s = 1: length(b)
for k = 1: length(c)
L(n,m,s,k)= a(n,m) +b(s)*a(n,m)*exp(c(k)*a(n,m));
end
end
end
end

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采纳的回答

Stephen23
Stephen23 2016-6-20
编辑：Stephen23 2016-6-20
tmp = bsxfun(@times,a,reshape(c,1,1,1,[]));
tmp = bsxfun(@times,a,exp(tmp));
tmp = bsxfun(@times,reshape(b,1,1,[]),tmp);
tmp = bsxfun(@plus,a,tmp);
Note that the floating point error propagates slightly differently, so isequal will be false.
  2 个评论
Amelos
Amelos 2016-6-21
编辑：Stephen23 2016-6-21
a =rand(2,2,3);
b = [ 1 2 3];
c = [1 2 3 4 5];
for n = 1: size(a,1)
for m = 1:size(a,2)
for s = 1: length(b)
for k = 1: length(c)
L(n,m,s,k)= a(n,m,s) +b(s)*a(n,m,s)*exp(c(k)*b(s));
end
end
end
end
Thanks for the answer! I m trying to understand the approach? What happens, if the dimension of the first matrix chances? See the example above.
Stephen23
Stephen23 2016-6-21
Well, you didn't just change the matrix dimensions, you also changed the operation by replacing the a(n,m) term inside the exp with a b(s) term. So lets do the same:
B = reshape(b,1,1,[]);
tmp = bsxfun(@times,B,reshape(c,1,1,1,[]));
tmp = bsxfun(@times,a,exp(tmp));
tmp = bsxfun(@times,B,tmp);
tmp = bsxfun(@plus,a,tmp);
and now compare some of the output values with your loop's output:
>> L(:,:,2,4)
ans =
4542.774 191.509
31.398 3096.897
>> tmp(:,:,2,4)
ans =
4542.774 191.509
31.398 3096.897

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