Hi everyone! I need help!

I need to get the probability of every element in a 3D matrix (Mat) which is 83x92x80

    % code
  %
[r, c, t] = size(Mat);
y = zeros(r,c,t);
p = zeros(r,c,t);
for i = 1:c;      

    for j = 1:r;          

        for k=1:t;

        y(j,i,k) = sum(Mat (:,:,[k]) == Mat(j,i,k));

        p(j,i,k) = y(j,i,k)/80;                
        end
    end
end

First, I am getting an error “Assignment has more non-singleton rhs dimensions than non-singleton subscripts”. Second, I am not quite sure if this is how I should do what I want to do. I really appreciate your help.

Many thanks

Endaw

[Bob Thompson]

Which line generates the error? Please write the line or copy paste the entire error message.

[Engdaw Chane]

Thank you Bob! The error occurring in the for loop. and the error is: "Assignment has more non-singleton rhs dimensions than non-singleton subscripts." I would happy to know another way of calculating the probability of each element in the third dimension.

Kindly, Engdaw

[Bob Thompson]

The error occurs because your code is trying to fit a three dimensional array, created by the sum() command, into a single element. If you're looking for the total summation, James suggested a decent solution.

Answer 1

[James Tursa]

Maybe adding another sum gets the result you want?

y(j,i,k) = sum(sum(Mat (:,:,[k]) == Mat(j,i,k)));

[Engdaw Chane]

Hi James, I tried it, and didn’t work. I just need to count how many times each element occurred in the third dimension, time. the result matrix can be just a two dimensional matrix. Thank you for helping me.

[James Tursa]

OK, at this point I think we need an example to understand what it is you want. Can you show us a sample small sized array, say 2x3x4, and show us this array and also show us the exact desired output for this array?

[Engdaw Chane]

My matrix has 83 rows, 92 columns and 80 months. What I need is the probability of each value along time (80 months). For example, this is how I calculated the probability of zero values for every row and column along time.

    % code
sum(matrix == 0, 3) ./ (size(matrix,3)); % here the result was just a matrix (83x92).
%

Now, I have to calculate the probability of all values.

For example if a value (at a specific row and column) occurred once in all 80 files, the probability for that value would be 1/80=0.0125. But this probability value changes if that specific value occurred multiple times in the 80 files.

thank you for your effort to save me!

Kindly, Chane

Answer 2

[Steven Lord]

Are you trying to compute the histogram of the elements in that array? If so take a look at the histogram function (or the histcounts function if you just need the counts without the picture.)

[Engdaw Chane]

Steven Lord

This is what I need to do; My matrix has 83 rows, 92 columns and 80 months. What I need is the probability of each value along time (80 months). For example, this is how I calculated the probability of zero values for every row and column along time.

    % code
sum(matrix == 0, 3) ./ (size(matrix,3)); % here the result was just a matrix (83x92).
%

Now, I have to calculate the probability of all values.

For example if a value (at a specific row and column) occurred once in all 80 files, the probability for that value would be 1/80=0.0125. But this probability value changes if that specific value occurred multiple times in the 80 files.

thank you for your effort to save me!

Kindly, Chane