nested for loop
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I can't figure out why the S matrix is computed to have the same value for every element - 0.7654. All other matrices (A through E) are computed correctly.
format short
x=[-0.9239, -0.6533, 0, 0.6533, 0.9239, 0.6533, 0, -0.6533]; %yj
y=[0, 0.6533, 0.9239, 0.6533, 0, -0.6533, -0.9239, -0.6533]; %yi
X=[-0.9239, -0.9239, -0.3827, 0.3827, 0.9239, 0.9239, 0.3827, -0.3827]; %Xj
Xone=[-0.9239, -0.3827, 0.3827, 0.9239, 0.9239, 0.3827, -0.3827, -0.9239];%Xj+1
Y=[-0.3827, 0.3827, 0.9239, 0.9239, 0.3827, -0.3827, -0.9239, -0.9239]; %Yj
Yone=[0.3827, 0.9239, 0.9239, 0.3827, -0.3827, -0.9239, -0.9239, -0.3827]; %Yj+1
phi=pi/180*[90, 45, 0, 315, 270, 225, 180, 135]; %phi
for i=1:8
for j=1:8
A(i,j)=-(x(i)-X(j)).*cos(phi(j)) - (y(i)-Y(j)).*sin(phi(j));
B(i,j)=(x(i)-X(j)).^2 + (y(i)-Y(j)).^2;
C(i,j)=sin(phi(i) - phi(j));
D(i,j)=(y(i)-Y(j)).*cos(phi(i)) - (x(i)-X(j)).*sin(phi(i));
E(i,j)=(x(i)-X(j)).*sin(phi(j)) - (y(i)-Y(j)).*cos(phi(j));
S(i,j)=((Xone(j)-X(j))^2 + (Yone(j)-Y(j))^2)^0.5;
end
end
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采纳的回答
Matt Fig
2011-5-8
The way you have defined S, it is independent of index i.
S(i,j) = ((Xone(j)-X(j))^2 + (Yone(j)-Y(j))^2)^0.5; % No i appears.
Therefore it will have the same value for every element of each column. Did you mean to have no dependence on index i?
4 个评论
Matt Fig
2011-5-8
You probably wouldn't notice the difference with such small arrays. But in general, this will make your code faster. This code can be vectorized with multiple calls to BSXFUN, but simply removing the inner FOR loop and dynamically pre-allocating the arrays would probably help speed-wise:
cosphi = cos(phi);
sinphi = sin(phi);
Sv = ((Xone-X).^2 + (Yone-Y).^2).^0.5;
for ii = 8:-1:1
A2(ii,:) = -(x(ii)-X).*cosphi - (y(ii)-Y).*sinphi;
B2(ii,:) = (x(ii)-X).^2 + (y(ii)-Y).^2;
C2(ii,:) = sin(phi(ii) - phi);
D2(ii,:) = (y(ii)-Y).*cos(phi(ii)) - (x(ii)-X).*sin(phi(ii));
E2(ii,:) = (x(ii)-X).*sinphi - (y(ii)-Y).*cosphi;
S2(ii,:) = Sv;
end
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