double summation without a loop

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Leo Simon
Leo Simon el 29 de Mzo. de 2015
Comentada: Jan el 30 de Mzo. de 2015
I have an anonymous function f that has two indices, i and t. i runs from 1 to n; t runs from 1 to m != n. For example:
f = @(delta,i,t) exp( - delta(i).*t );
delta = rand(1:n);
Obviously, I can sum f with respect to either t or i,
sum(f(delta,1:n,t))
sum(f(delta,i,1:m))
But I'd like to compute, in one step, sum_i=1^n (sum_t=1^m f(delta,i,t)) , without writing a loop. Is this possible? thanks very much for any advice!

Respuesta aceptada

Jan
Jan el 29 de Mzo. de 2015
f = @(delta,i,t) exp(bsxfun(@times, -delta(i), t.'))
result = sum(reshape(f(delta, 1:n, 1:m), 1, []))
  3 comentarios
Stephen23
Stephen23 el 30 de Mzo. de 2015
Editada: Stephen23 el 30 de Mzo. de 2015
In general operations with a period character . are elementwise operations, whereas the operations without are standard linear algebra operations.
In this case:
Jan Simon used the non-conjugate transpose in their answer, and this is the one you should use too.
Jan
Jan el 30 de Mzo. de 2015
@Stephen: In their answer? I'm aware that the name Jan can be male and female and Simon is a common first name also, but we are at least only one person. So I'd prefer his ;-)

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