Rearranging matrices

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Donald
Donald 2011-5-5
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!
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Andrei Bobrov
Andrei Bobrov 2011-5-5
as suggested by Sean de
data_new = permute(reshape(data,[6,28,6]),[1 3 2])
Donald
Donald 2011-5-5
I'm pretty new to MATLAB, but I'll check that out. Thanks!

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回答（3 个）

Andrei Bobrov
Andrei Bobrov 2011-5-5
more variant
data_new = reshape(permute(reshape(data,[6,28,6]),[1 3 2]),6,[]);
Teja Muppirala
Teja Muppirala 2011-5-5
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
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Donald
Donald 2011-5-5
thanks!
Teja Muppirala
Teja Muppirala 2011-5-5
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.

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Matt Fig
Matt Fig 2011-5-5
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,:)

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