reshaping 2D matrix into 3D - specific ordering

Hi,
I'd like to ask for your help in the following problem:
I have an n by m 2D matrix which is formed by concatenating a number of k by m matrices, where k < n and n/k is a positive integer. I want to construct a three dimensional matrix that stores each k by m matrix in a different layer in the third dimension. For example:
a =
0.6443 0.1948 0.5949 0.7303 0.0377
0.3786 0.2259 0.2622 0.4886 0.8852
0.8116 0.1707 0.6028 0.5785 0.9133
0.5328 0.2277 0.7112 0.2373 0.7962
0.3507 0.4357 0.2217 0.4588 0.0987
0.9390 0.3111 0.1174 0.9631 0.2619
0.8759 0.9234 0.2967 0.5468 0.3354
0.5502 0.4302 0.3188 0.5211 0.6797
0.6225 0.1848 0.4242 0.2316 0.1366
0.5870 0.9049 0.5079 0.4889 0.7212
0.2077 0.9797 0.0855 0.6241 0.1068
0.3012 0.4389 0.2625 0.6791 0.6538
0.4709 0.1111 0.8010 0.3955 0.4942
0.2305 0.2581 0.0292 0.3674 0.7791
0.8443 0.4087 0.9289 0.9880 0.7150
I want b to be
b
b(:,:,1) =
0.6443 0.1948 0.5949 0.7303 0.0377
0.3786 0.2259 0.2622 0.4886 0.8852
0.8116 0.1707 0.6028 0.5785 0.9133
0.5328 0.2277 0.7112 0.2373 0.7962
0.3507 0.4357 0.2217 0.4588 0.0987
b(:,:,2) =
0.9390 0.3111 0.1174 0.9631 0.2619
0.8759 0.9234 0.2967 0.5468 0.3354
0.5502 0.4302 0.3188 0.5211 0.6797
0.6225 0.1848 0.4242 0.2316 0.1366
0.5870 0.9049 0.5079 0.4889 0.7212
b(:,:,3) =
0.2077 0.9797 0.0855 0.6241 0.1068
0.3012 0.4389 0.2625 0.6791 0.6538
0.4709 0.1111 0.8010 0.3955 0.4942
0.2305 0.2581 0.0292 0.3674 0.7791
0.8443 0.4087 0.9289 0.9880 0.7150
What's the most efficient way to do this (preferably without the need for a loop)? The function 'reshape' picks elements column-wise, so it is not a viable option in this case.
Thank you,
Paul

 Akzeptierte Antwort

Oleg Komarov
Oleg Komarov am 25 Apr. 2012

11 Stimmen

[r,c] = size(a);
nlay = 3;
out = permute(reshape(a',[c,r/nlay,nlay]),[2,1,3]);
Yes, MATLAB does 'think' along rows. Then, transpose and reshape. Then, you need to re-'transpose' again but you have a 3D array thus use permute (nD general 'transpose').

5 Kommentare

Pal
Pal am 25 Apr. 2012
Thanks!
Erwin Schoo
Erwin Schoo am 23 Nov. 2013
This was a life saver for me. Thanks!
Dev
Dev am 16 Jul. 2014
Bearbeitet: Dev am 16 Jul. 2014
I was looking exactly for this.
Parreche
Parreche am 3 Sep. 2014
thanks!!
When I do this, I get
Error using reshape
Size arguments must be integer scalars.
Is there any way around this? I'm trying to un-reshape a 2D image back to 3D image with a dimension of 144. Here is the original reshaping:
[r c d] = size(data); % Originally a 340x740x144 array
img = reshape(data,r*c,d); % This gives me a 251600x144 array
Now I want to go backwards, I did some editing to the image, and now I want to make it 3D again, with 144 dimensions.

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Pal
am 25 Apr. 2012

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am 23 Apr. 2015

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