How do I store matrices into a cell array in the correct order?

I'm trying to store a set of matrices that are generated using a for loop. The structure of the code is

for k = 1:N % loops through some folders - a few commands go in here for sorting the folders
    for kk = n:-1:1 % loops through the files in each folder in reverse - the files are csvs that contain two columns of data. I then use the find peaks function on this data and for each file store a matrix
        matrix = [loc height]; % this stores the position of each peak and its height - for each file this may have a different number of rows as they do not all have the same number of peaks
        cellarray{kk} = matrix; % Should store each matrix in a cell array
    end
end

When I store the cell array, for some reason the elements are stored in reverse order. The matrix from the first file that the code loops through is the last cell in the cell array. I can't think of a good reason why this should be the case, other than maybe it is to do with the fact that I am looping through the files in reverse. If I look at matrix, matrix(1) is a 6x2 matrix and matrix(end) is a 1x2 matrix, but in the cell array, the first cell is 1x2 and the last cell is 6x2. Why is this the case and what can I do about it? If I reverse the order of the kk iterations, it loops through matrix in the reverse order but store cellarray in the same way.

[Rik]

You are not showing your complete code, which makes it hard to come up with definitive answers. It looks like the kk counter determines which file you analyze and which cell position the results are stored in. That means that reversing the loop doesn't change the ordering of the cell array, just the order in which it is filled. If you want to store them in reverse order you will need something like this:

A=1:5;
B=zeros(size(A));
for n=numel(A):-1:1
    B(numel(A)-n+1)=A(n);
end

[Jan]

@JJH: This piece of code overwrites the elements of cellarray in each iteration of the outer loop. Is this intented?

Answer 1

[Stephen23]

"When I store the cell array, for some reason the elements are stored in reverse order."

Not at all. They are stored in the exact order that you specify using indexing, as I explained earlier:

https://www.mathworks.com/matlabcentral/answers/434237-how-do-i-flip-the-order-of-cells-in-a-cell-array#comment_647414

which has nothing to do with which one you define first or last. Index 1 is always index 1, no matter when you assign to it. Lets go through two simple examples step by step, first in descending order, like your code:

C = nan(1,5);
for k = 5:-1:1 % descending order
    C(k) = k^2;
end

The loop iterations are:

1. assign 5^2 to C(5)
2. assign 4^2 to C(4)
3. assign 3^2 to C(3)
4. assign 2^2 to C(2)
5. assign 1^2 to C(1)

giving the vector [1,4,9,16,25]. Now lets do it in asending order:

D = nan(1,5);
for k = 1:5 % ascending order
    D(k) = k^2;
end

The loop iterations are:

1. assign 1^2 to D(1)
2. assign 2^2 to D(2)
3. assign 3^2 to D(3)
4. assign 4^2 to D(4)
5. assign 5^2 to D(5)

giving the vector [1,4,9,16,25]. Because our iterations are totally independent of one another, their order of calculation is totally irrelevant. Whether ascending or descending in the loop, we get the same output vector.

Of course if you want to change the indices then you have to change the indices (not the order of the loop iterations, which makes no difference):

C = cell(1,n);
for kk = 1:n
    C{n-kk+1} = your data
end

Or perhaps simpler to reverse afterwards:

C = cell(1,n);
for kk = 1:n
    C{kk} = your data
end
C(end:-1:1) = C(:)

which is exactly what I showed you one day ago:

https://www.mathworks.com/matlabcentral/answers/434237-how-do-i-flip-the-order-of-cells-in-a-cell-array#answer_350894

[JJH]

Ok, I think I see what you mean now. It seems to be working, thanks for your help.