Rearranging matrices

Hi! I am working with a 168x6 matrix that is actually just 28 monthly 6x6 matrices stacked on top of each other. To get the data in a way that is easier for me to work with, I would like to rearrange the matrices so that each 6x6 matrix is placed side by side in a row rather than a column. A simple transpose does not work because the data are in one large matrix and the months are by no means separated. Any recommendations? Thanks!

[Sean de Wolski]

Are you sure it wouldn't be more useful to have them stacked along the third dimension? Just a thought..

[Andrei Bobrov]

as suggested by Sean de

data_new = permute(reshape(data,[6,28,6]),[1 3 2])

[Donald]

I'm pretty new to MATLAB, but I'll check that out. Thanks!

Answer 1

[Andrei Bobrov]

more variant

data_new = reshape(permute(reshape(data,[6,28,6]),[1 3 2]),6,[]);

Answer 2

[Teja Muppirala]

If your original matrix is called "data"

data_new = zeros(6,168);

for n = 0:(168/6)-1
    data_new(:,6*n + (1:6)) = data(6*n + (1:6),:);
end

[Donald]

thanks!

[Teja Muppirala]

What I wrote is the most obvious, simple solution for this problem. But if you need to do this calculation hundreds of thousands of times, or repeatedly on very large datasets (say, 6000000x6), I would actually recommend using andrei's solution because it is actually significantly faster.

Answer 3

[Matt Fig]

I agree with Sean de that indexing into your arrays would be easier if they were stacked in the third dimension.

B = permute(reshape(permute(A,[2 1 3]),6,6,28),[2,1,3]);

Now your first array is:

B(:,:,1)

and the second is:

B(:,:,2)

etc.

Also, you might consider cell arrays:

C = permute(reshape(permute(A,[2 1 3]),6,6,28),[2,1,3]);
C = mat2cell(B,6,6,ones(1,28));

Now your first array is:

C{1}

and your second array is:

C{2}

etc. This makes indexing (getting at a certain array) a lot easier!

If you are unfamiliar with cell arrays, you get to the 3rd row of the 5th array like this:

C{5}(3,:)

[Matt Fig]

Mental gymnastics this morning...