better method for evaluating matrix

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JaeSung Choi
JaeSung Choi am 21 Nov. 2017
Bearbeitet: Stephen23 am 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);

Akzeptierte Antwort

Stephen23
Stephen23 am 21 Nov. 2017
Bearbeitet: Stephen23 am 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))

Weitere Antworten (1)

Andrei Bobrov
Andrei Bobrov am 21 Nov. 2017
Bearbeitet: Andrei Bobrov am 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 Kommentar
Stephen23
Stephen23 am 21 Nov. 2017
Bearbeitet: Stephen23 am 21 Nov. 2017
Note that the output is transposed compared to the code given in the question.

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