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I want to divide an image data into blocks, for example if I have X matrix of 4*4, I want the result to be a matrix M of 2*2*2*2 where M(1,1,:,:)=X(1:2,1:2) and M(1,2,:,:)=X(1:2,3:4) and etc.

I found a way to divide it into a cell array using mat2cell but cell arrays seem not very supported in matlab, I search and ask on SO just to do things that can be done easily with ordinary matrices(and I get answers suggesting not to use cell arrays at all).

I searched the net and SO thoroughly , there are many many results all of them either solve a particular problem(like finding a solution to an equation, which I don't want here) or end up at suggesting mat2cell.

The closest result I found though is using reshape, I tried reshape(X,[2 2 2 2]) and got C=2*2*2*2 matrix but C(1,1,:,:) seems to be the first row of X not the first block and I can't figure out what to do next.

Someone suggested permute(C,[1 2 3 4]) but I still don't get the result I want.

I'm still beginner to matlab so forgive me if my question is simple, also I don't want for-loops, I want vector code, so What to do next after reshape ?

Edit

I tried the answer of that duplicate question and still I don't get the result I want, I tried these: 

A=ones(40,40);
A(1:10,1:10)=32*ones(10,10);
A(11:20,1:10)=zeros(10,10);    
T = permute(reshape(A, size(A, 1), 10, []), [2 1 3]);  
T = permute(reshape(T, 10, 10, [], size(T, 3)), [2 1 3 4]);

T was 10*10*4*4, exactly the size I want(40*40 into a matrix of submatrices each of 4*4) but the values are wrong, T(1,1,:,:)~=A(1:4,1:4)

What Am I doing wrong here ?

niceman
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  • Specifically, see second variant of [this](http://stackoverflow.com/a/20337173/2586922) answer, with `m=2; n=2;` in your example – Luis Mendo May 18 '15 at 11:09
  • hmmm yes I can see, although I wouldn't know that's a duplicate except If I read the answers because of the question title – niceman May 18 '15 at 11:10
  • are you sure you do not need - `M(:,:,1,1) = X(1:2,1:2), M(:,:,2,1) = X(3:4,1:2), M(:,:,2,1) = X(1:2,3:4) and M(:,:,2,2) = X(3:4,3:4)` instead of what you have written ? – Nishant May 18 '15 at 11:13
  • @Nishant No I don't need that, I want `M(1,1,:,:)` not `M(:,:,1,1)` – niceman May 18 '15 at 11:17
  • @niceman You may need to `permute` the result of A.Donda's answer then – Luis Mendo May 18 '15 at 11:31
  • try `M= permute(reshape(X',[2,2,2,2]),[4,2,3,1]);`. I have checked it for the initial case – Nishant May 18 '15 at 12:09
  • @Nishant thanks, it worked, hmmmm can you post it as an answer(for future benefits) – niceman May 18 '15 at 15:18
  • I cannot post an answer as this question is marked as duplicate. I suggest you add this answer to your question. – Nishant May 18 '15 at 15:26
  • @Nishant - I've reopened the question. Please post your answer. – rayryeng May 18 '15 at 17:22
  • sorry for interupting.. But I was intrigued by your initial approach with `mat2cell` and saw no reason why no. You almost had it, only missing line is `cat(3,M{:})` with M beeing the result of `M=mat2cell(X,[2 2],[2 2])` – marco wassmer May 18 '15 at 17:44
  • @marcowassmer I believe mat2cell will be slower. Also, OP needs a matrix at the end. – Nishant May 19 '15 at 05:07
  • @Nishant did not know speed was an issue, I just found it to be kind of neat all in two lines... But I belive with M{:} will produce matrices and the end product be therefore a matrix – marco wassmer May 19 '15 at 06:00
  • @marcowassmer yes the result is a matrix but speed is an issue, good suggestion anyway :) – niceman May 19 '15 at 16:21

1 Answers1

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For the initial case (4x4 into 2x2x2x2x2) use

M= permute(reshape(X',[2,2,2,2]),[4,2,3,1]);

For a general case where size(X) = a x b) and size(M) = c x d x e x f such that a*b = c*d*e*f, for any natural number n, use

M= permute(reshape(X',[c,d,e,f]),[4,2,3,1]);
Nishant
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