matrix dimension error: need some hint!

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!

Answer 1

[Walter Roberson]

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.

[Image Analyst]

Also note that Xt(2:end)-Xt(1:end-1) is the same as diff(Xt).

Answer 2

[Charles Martineau]

Superb! Thanks a lot!!!