matrix dimension error: need some hint!
显示 更早的评论
Hi,
I have a vector "Xt" where Xt dimensions are 12588 X 1. From this vector I apply this code:
S=sum((abs(Xt(2:end)-Xt(1:end-1))).^2);
which result in one number where S is a 1by1.
Now my objective is to construct a vector S (Nx1) where each value in S depends on J and K -- for instance:
S=sum((abs(Xt(j:end)-Xt(k:end-1))).^2);
At first I thought of the following (I don't want J do exceed 126):
for j=2:126 k=1:125,
S=sum((abs(Xt(j:end)-Xt(1:end-k))).^2);
end;
I got this: "Error using - Matrix dimensions must agree."
Any insight at how I can correct this matrix issue?
Thank you!
采纳的回答
Walter Roberson
2012-5-26
Your line
for j=2:126 k=1:125,
is equivalent to
for j=2:126;
k=1:125
That is, assign k the vector 1:125 each time through the "for j" loop.
I get the impression that you may have been trying to loop over j and k. If so then you need "for" statements for each of the loops.
If you were to correct that, then you have the problem that Xt(j:end) is not going to be the same length as Xt(1:end-k) so you will not be able to subtract the two vectors.
Also, you overwrite "S" each time through the loop, which you probably do not want to do.
Perhaps you want something like
for j=2:126
S(j-1) = sum((abs(Xt(j:end)-Xt(1:end-j+1))).^2);
end
If that is the case, then
S = fliplr( cumsum( fliplr( (Xt(2:end)-Xt(1:end-1)).^2 ) ) );
Note that unless you are working with complex numbers, you do not need to abs() numbers before squaring them.
1 个评论
Image Analyst
2012-5-26
Also note that Xt(2:end)-Xt(1:end-1) is the same as diff(Xt).
更多回答（1 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Creating and Concatenating Matrices 的更多信息
2012-5-26
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
