why embedded matlab function failed to determine the size of expression
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Hi all,
u is an input from the output of continous state.so, for each time (1 x 3) matrix will come out, so
u matrix = 1 x 3
Mean = 1 x 450
is a mean of samples.This input from workspace which consist 150 times with 3 variables so its matrix is (1 x 450). but we only need (1 x 3) matrix to evaluate for each time
Std = 1 x 450
is a standard deviation of samples.Same description as above(mean)
Xnew = 1x3
is an outpyt from the equation Xnew = u-mean/std
Example;
for time 1
Xnew = u-Mean(1,1:3)
time 2
Xnew = u-Mean(1,4:6);
untill time 150
i'm trying to do loop in embedded matrix function as below command:
function Xnew = stats(u,Mean, Std)
%#eml
for j=1:150,
Xnew = (u-Mean(1,j:j+2))./Std(1,j:j+2);
end
unfortunately i received this error,
"Could not determine the size of this expression"
I'm doing looping because i want to pick the submatrix that will be used in the equation.
What should i change above command so as i can get the result as i want? anybody have idea?
1 Kommentar
Jan
am 18 Jul. 2012
I have problems to determine the size of u also. Are u and Mean, Std and XNew different variables? Then please edit the question and brush up the formatting.
Does Matlab have a chance to know the size of u?
Is there a reason to overwrite XNew in each iteration of the for j loop?
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Fred Smith
am 18 Jul. 2012
In more recent versions your code should just work. However, in older versions the size of this pattern was not understood:
j:j+2
Simply rewrite it as:
j + (0:2)
HTH,
Fred
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