better method for evaluating matrix

1 vue (au cours des 30 derniers jours)
JaeSung Choi
JaeSung Choi le 21 Nov 2017
Modifié(e) : Stephen23 le 21 Nov 2017
For given vector such as following, I want to make square matrix s such that
%constant, original form
N = 500;
x = 2*pi*linspace(0,1,N);
for i = 1:N
for j = 1:N
s(i,j) = sin(x(i)-x(j))
end
end
But it was too slow, so recently I edited it to following, but it's still too slow!!
Can anybody please help me??
%constant
N = 500;
x = 2*pi*linspace(0,1,N);
for i = 1: N;
s(:,i) = x-x(i);
end
s=sin(s);

Réponse acceptée

Stephen23
Stephen23 le 21 Nov 2017
Modifié(e) : Stephen23 le 21 Nov 2017
You really need to learn how to write vectorized code. Solving every task using lots of ugly loops is not an efficient way to write MATLAB code. Try this:
N = 500;
vec = 2*pi*linspace(0,1,N);
mat = sin(bsxfun(@minus,vec(:),vec))

Plus de réponses (1)

Andrei Bobrov
Andrei Bobrov le 21 Nov 2017
Modifié(e) : Andrei Bobrov le 21 Nov 2017
N = 500;
x = 2*pi*linspace(0,1,N);
s = sin(x(:)' - x(:));
for old versions of MATLAB:
s = sin( bsxfun(@minus, x(:)',x(:)) );
  1 commentaire
Stephen23
Stephen23 le 21 Nov 2017
Modifié(e) : Stephen23 le 21 Nov 2017
Note that the output is transposed compared to the code given in the question.

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