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How to recover a square matrix from its upper triangular cell part?

9 vues (au cours des 30 derniers jours)
Xh Du
Xh Du le 27 Juil 2017
Commenté : Xh Du le 28 Juil 2017
Hi all,
I'd like to compute the upper triangular part only for a symmetric matrix in a block by block way. After some operations I got this:
K>> tempBlk{:}
ans =
[6x6 double] [6x5 double] [6x5 double] [6x5 double]
[] [5x5 double] [5x5 double] [5x5 double]
[] [] [5x5 double] [5x5 double]
[] [] [] [5x5 double]
The cell blocks in the diagonal line here are full, symmetric, and the non-diagonal cell blocks are full, non-symmetric. The entire symmetric matrix result should be 21 by 21, which has the length and width of 6+5+5+5. I'd like to fill the empty cells (these [ ]) with zeros, then use cell2mat to transform it back to scalar matrix, then use triu to take only the upper triangular part.
Any idea of how to do this? Thanks!

Réponse acceptée

Matt J
Matt J le 27 Juil 2017
Modifié(e) : Matt J le 27 Juil 2017
There's no reason to fill the [] cells with zeros. You can fill them with anything, since they will eventually be overwritten with zeros by triu. I would therefore do as follows:
idx=cellfun('isempty', yourCellArray);
C=cellfun(@transpose,yourCellArray.','uni',0);
yourCellArray(idx)=C(idx);
result = triu(cell2mat(yourCellArray))
  5 commentaires
Matt J
Matt J le 27 Juil 2017
Probably, but it will amount to a 25% saving at best. It would be better if, instead of cell2mat, you just allocated the final matrix at the very beginning and wrote into the blocks.
Xh Du
Xh Du le 28 Juil 2017
Thank you, this is very helpful!

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