In a for loop matrix dimension changes?

조회 수: 2 (최근 30일)
Catarina
Catarina 2014년 12월 6일
댓글: Catarina 2014년 12월 9일
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 2014년 12월 6일
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.
  댓글 수: 3
Roger Stafford
Roger Stafford 2014년 12월 9일
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 2014년 12월 9일
Ok I've figured it out now! Thank you for your help! :)

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