How to create a block diagonal matrix from the slices of a 3D matrix

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I have a large 3D matrix (K) composed of 200 slices , and I want to create a block diagonal matrix from the slices of it without looping. For example
K(:,:,1) =[1 2;3 4]
K(:,:,2) =[5 6;7 8]
K(:,:,3) =[9 10;11 12]
.
.
.
K(:,:,200) =[1 10;13 15]
if we assume the number of slice is 3, the output should be
1 2 0 0 0 0
3 4 0 0 0 0
0 0 5 6 0 0
0 0 7 8 0 0
0 0 0 0 9 10
0 0 0 0 11 12
But how the block diagonal matrix created from all the 200 slices without looping?

Risposta accettata

Azzi Abdelmalek
Azzi Abdelmalek il 2 Ott 2014
k(:,:,1) =[1 2;3 4]
k(:,:,2) =[5 6;7 8]
k(:,:,3) =[9 10;11 12]
q=num2cell(k,[1,2])
blkdiag(q{:})
  4 Commenti
Azzi Abdelmalek
Azzi Abdelmalek il 2 Ott 2014

هذه إحدى التقنيات تعلمتها في هذا المنتدى

Matt J
Matt J il 2 Ott 2014
Note that this still uses a for-loop inside num2cell.m, though how much that hurts you, I don't know.
Note also that my 2nd answer here was the same as this, except that it returns the result in sparse form, which is normally more desirable for block diagonal matrices with small blocks.

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Più risposte (3)

Matt J
Matt J il 2 Ott 2014
Modificato: Matt J il 2 Ott 2014
I doubt that avoiding loops leads to the fastest code, but the following does avoid them,
[m,n,p]=size(K);
BlockMatrix=kron(speye(p), ones(m,n));
BlockMatrix(logical(BlockMatrix))=K(:);

Matt J
Matt J il 2 Ott 2014
Modificato: Matt J il 2 Ott 2014
This doesn't avoid a loop, even if it looks like it does, but it might be the fastest way.
Kcell = cellfun(@sparse, num2cell(K,[1,2]), 'uni',0 );
BlockMatrix=blkdiag(Kcell{:});

Azzi Abdelmalek
Azzi Abdelmalek il 2 Ott 2014
k(:,:,1) =[1 2;3 4]
k(:,:,2) =[5 6;7 8]
k(:,:,3) =[9 10;11 12]
out=[];
for ii=1:size(k,3)
out=blkdiag(out,k(:,:,ii))
end

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