In a for loop matrix dimension changes?

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Catarina
Catarina 2014-12-6
评论： Catarina 2014-12-9
I'm having a problem. Here is my code:
UFR = 0.042;
alpha = 0.1;
N = 20;
%Wilson function
for p = 1:N
for q = 1:N
W = SWfunction(p,q,alpha,UFR);
end
end
w = inv(W);
for t = 1:s
for h = 1:N
M(:,h) = exp(-(h)*R(t,h));
MU(:,h) = exp(-UFR*h);
Z(:,h) = zeros;
m = transpose(M);
mu = transpose(MU);
z = transpose(Z);
z(:,t) = w.*(m-mu);
end
end
When I run my code matlab tells me matrix dimensions do not agree when calculating z(:,t). But here's what confuses me, when I run the code without that line the matrix dimensions are fine for multiplication (20x20 with 20x1) but when I have that line it changes the variables m and mu to 1x1 matrices... I need z to be a matrix that performs that calculation for all t. Can someone help me?

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Roger Stafford
Roger Stafford 2014-12-6
If 'SWfunction' produces a scalar value, then 'w' will be a scalar, that is 1 x 1, since it is being overwritten 400 times in your first nested loops. That would produce an error at the line
z(:,t) = w.*(m-mu);
You need to use p and q to index 'w' values, as in w(p,q), so as to produce a 20 x 20 matrix.
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Roger Stafford
Roger Stafford 2014-12-9
In the code you show in the above comment the quantity 'w' is still a 1 x 1 scalar. You need W(p,q) instead of W in the line
W = SWfunction(p,q,alpha,UFR);
A second problem is that in the nested for loops that create M and MU you are taking the transpose of arrays that are only partially computed. That makes no sense. You need to be creating the 'm', 'mmu' and 'z' arrays in a later action after M and MU are fully computed.
Also the line
z(:,t) = w(p,q)*mmu(:,t);
makes no sense. The p and q indices have long since become obsolete since they refer to indices in the first set of for loops.
Catarina
Catarina 2014-12-9
Ok I've figured it out now! Thank you for your help! :)

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