speed up 'for' loops

4 vues (au cours des 30 derniers jours)
Coo Boo
Coo Boo le 7 Sep 2012
Hi,
How can I speed up following 'for' loops? Help me please.
P=200;
N=40000000;
y(1:P)=3;
% a: P X 1 matrix (vector)
% Z: P X 1 matrix (vector)
% x: N X 1 matrix (vector)
%%%%%Loops
y(P+1:N)=0;
for i=P+1:N
for j=1:P
y(i)=y(i)-a(j)*x(i-j);
end
end
for i=1:N
for j=1:P
f(i,j)=Z(j)^(i-1);
end
end
Thanks in advance.

Réponse acceptée

Jan
Jan le 7 Sep 2012
Modifié(e) : Jan le 7 Sep 2012
P = 200;
N = 40000000;
a = rand(P, 1);
Z = rand(P, 1);
x = rand(N, 1);
y(P+1:N) = 0; % This one at first! -> pre-allocation
y(1:P) = 3;
at = transpose(a);
for i=P+1:N
y(i) = y(i) - at * x(i-1:-1:i-P); % Dot-product of vectors => SUM
end
f = ones(P, N);
for i = 2:N
f(:, i) = f(:, i - 1) .* Z;
end
f = transpose(f);
  5 commentaires
Jan
Jan le 7 Sep 2012
Modifié(e) : Jan le 7 Sep 2012
Thanks, Coo Boo, fixed now. Unfortunately I cannot test it, because I do not have a Matlab version installed on my current computer. But you are cordially invited to debug it.
Coo Boo
Coo Boo le 7 Sep 2012
Thank you very much

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Plus de réponses (1)

Azzi Abdelmalek
Azzi Abdelmalek le 7 Sep 2012
for the second loop
c=repmat(Z',N,1)
f=bsxfun(@power,c,[0:N-1]')
  4 commentaires
Coo Boo
Coo Boo le 7 Sep 2012
Thank you very much
Matt Fig
Matt Fig le 7 Sep 2012
No need for REPMAT or [].
f = bsxfun(@power,Z.',(0:N-1).');

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