index exceed matrix dimensions

hello folks, I'm trying to run 2 different dimensional vectors by using weighted regions with a chosen factor. can I get some expert opinion on identify the error,please. your help is gratefully appreciated.Many thanks in advance.

   x=train_data; % 882 x2 double
  z=test_data;  %882 x 8 double
  distance_max=[];

for h=1:size(x)    
 for kb=1:size(z)

        kk=x(:,h);
        gg=z(:,kb); <-- this is where the error appeared

weightsum=0;
ww  = [0, 1,1,1,1,1,0,0,1,1,1,1,1,0,1,1,2,1,2,1,1,2,4,4,0,4,4,2,1,1,1,1,1,1,1,0,1,2,2,2,1,0,0,0,1,1,1,0,0];   
length=18;
region = size(x,1)/length; 

    for e=1:region

      e_start = (e -1) * length+1 ;
      e_end = e_start + length-1 ; 

     xpart=kk(e_start:e_end);   
     zpart = gg (e_start:e_end);

   D1 = pdist2(xpart',zpart','chisq')*ww(e);
   distance_max=[distance_max,D1];
  weightsum=weightsum+D1;

    end
  results(h,k)={weightsum};    
  end

 end

[Adam]

Presumably you got a line number with that error that tells you the line which causes the error?

It is a lot more difficult for us to look through a whole block of code hunting for an error when the error message should just point straight to it!

What is y by the way?

[seprienna]

this is the line

Index exceeds matrix dimensions.
 Error in weight (line 13)
      gg=z(:,kb);

I have edited it in the code.

There is no 'y', its only 'x'-training data and 'z'-testing data

[Adam]

There is a

 for kb=1:size(y)

line in your code though so when I tried to run it with random matrices of the correct size in place of your x and z that threw an error.

Should that just be size(z) then instead of size(y)?

[seprienna]

million apologies for the confusion adam. I was using the y for a different data but in this experiment, i'm using 'x' and 'z'. Can you kindly point out how can i avoid the dimension error please.

[Rick Rosson]

Two lines above the line that caused the error:

   for kb=1:size(y)

Answer 1

[Rick Rosson]

Please try changing the line

   for kb=1:size(y)

to the following:

   for kb=1:size(z)

Also, please correct the issue with size described in the answer provided by Adam.

[seprienna]

Hello Rick,

I'm aware with the size problem. I have followed your advise, but it looks like I'm not getting what i wanted. If you could look above for my comment to Adam, maybe you can enlighten my issue. your opinion is greatly appreciated.

[Rick Rosson]

I think you want to loop over the number of columns in each matrix. So please try the following:

   for h=1:size(x,2)    
       for kb=1:size(z,2)

[seprienna]

Rick,thanks for the tips. that is the result i'm looking for. thanks again.

Answer 2

[Adam]

size(x)

will return a length d vector where d is the dimension of x. In your case that will be [882 2] and likewise size(y) will return [882 8].

If you want the first dimension you can use

size( x, 1 )

to give 882, or

size( x, 2 )

if you want the second dimension only. You can use

length(x)

if you just want the longest dimension but don't know what the index of that dimension is, but I don't recommend using length for this purpose if you do know which dimension you want.

That may not be the only problem, but it will definitely cause problems.

[seprienna]

hello Adam, many thanks for the helpful information.

i need both dimensions as the are 2 images in the (x) and (z) are the 8 testing images.

i need to calculate the chi-square distance between (x) and (z).

Initially I used size(x,1) to calculate 1 image only, but with 2 images, i wish to do it in a loop, I read from the MATLAB guide that it is possible, but I'm not sure where I made the mistake.

[Pierre Benoit]

Well from what your code do, it seems that you need to use

size(x,2)

and the same for z.

[seprienna]

it works!!!thanks you very much