How to avoid nested for loop to make the code faster?

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Stef
Stef 2019 年 3 月 14 日
回答済み: Doug Mercer 2019 年 3 月 14 日
Hi everyone!
Do you have some suggestions to improve the following code?
As it is now, it is very slow…
B_2 = zeros(d,d);
for m = 1:M
for n = m+1 : 2*N
B_2 = B_2 + 4*N^(-1)* A(n) * B(n-m);
end
end
where A(n) and B(n) are 2x2 real, symmetric and full rank, matrices for all possible value of n.
Thanks in advance!
  6 件のコメント
Stef
Stef 2019 年 3 月 14 日
@Adam The matrices A(n) and B(n) are 2x2 matrices for every possible value of n (e.g. A(1) is a 2x2 matrix). In particular, for each n, I evaluate A(n) and B(n) as outer multiplication of two pre-stored, 2x1 column vectors, i.e. A(n) = u(n)*v(n).'
Stef
Stef 2019 年 3 月 14 日
d = 2,
N = 10^5 (could be also 10^6),
M is of the order of sqrt(M), but it can change...

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採用された回答

Doug Mercer
Doug Mercer 2019 年 3 月 14 日
If you can generate A and B in a vectorized manner (i.e., for inputs x=[x_1, x_2, ..., x_k] the function A(x) returns a 2x2xk 3D array where a(:, :, i) is the 2x2 array corresponding to x_i) then the following would vectorize the inner loop.
function B_2 = stef()
d = 2;
N = 10^5;
M = sqrt(N);
B_2 = zeros(d,d);
for m = 1:M
n_iter = m+1:2*N;
B_2 = B_2 + 4*N^(-1)*sum(A(n_iter).*B(n_iter - m), 3);
end
function b = B(x)
b = rand(2, 2, length(x));
function a = A(x)
a = rand(2, 2, length(x));

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