Rearrange 3 small matrices to a bigger one with a certain order: taking first n-rows of each of the smaller matrices after each other

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Tim 2014-1-3

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编辑： Andrei Bobrov 2014-1-6

采纳的回答： Azzi Abdelmalek

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I have three smaller Matrices that I want to transform to one bigger one with the following order:

>> A1 =

A2 =

They should be transformed to one Matrix in this order:

B =

So always take the first n-rows of each matrix (here n=2, in the actual case 'n' ist way higher), then take all the second "packages" and so on (In this case: Take first two lines of each Matrix, then take lines 3-4 of each matrix, then lines 5-6 of each matrix and so on)

Is there a nice and quick solution that does not require loops? Maybe using reshape or some functions to swap lines or using indices or so..?

Thanks in advance again!

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Answer 1

Azzi Abdelmalek 2014-1-3

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编辑：Azzi Abdelmalek 2014-1-3

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m=size(A1,2);
n=2
A11=permute(reshape(A1',m,n,[]),[2 1 3]);
A22=permute(reshape(A2',m,n,[]),[2 1 3]);
A33=permute(reshape(A3',m,n,[]),[2 1 3]);
A=permute([A11;A22;A33],[2 1 3]);
B=A(:,:)'

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Tim 2014-1-6

perfect, thanks!

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Answer 2

Andrei Bobrov 2014-1-6

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编辑：Andrei Bobrov 2014-1-6

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a1 = cat(1,A1,A2,A3);
t = rem(0:size(a1,1)-1,4)+1 <= 2;
out = [a1(t,:);a1(~t,:)];

or in one row

out = reshape(permute(reshape(cat(1,A1,A2,A3),2,2,3,[]),[1 3 2 4]),[],2);

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Jos (10584) 2014-1-6

This only works for n is 2 ...

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Answer 3

Jos (10584) 2014-1-6

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You can use some clever indexing:

A1 = cumsum(ones(4,2)), A2 = A1 + 4, A3 = A1 + 8 % as above
n = 2 ;
% engine
m = size(A1,1) ; % all the same
  % build a clever sorting matrix IX
IX = repmat([1+floor((0:m-1)/n) ; zeros(1,m) ; 1:m].', 3, 1) 
IX(:,2) = 1+floor((0:(3*m)-1)/m)
  % col 1 : selection order based on n
  % col 2 : which matrix (A1, A2 or A3),
  % col 3 : which element inside matrix
% sort them nicely
  [IXsorted,SI] = sortrows(IX)
  B = [A1 ; A2 ; A3] 
  B = B(SI,:)