Can someone do this calculation without for loops?

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

Answer 1

[Stephen23]

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.

[Amelos]

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]

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