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

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Xh Du 2017-7-27

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评论： Xh Du 2017-7-28

采纳的回答： Matt J

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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!

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采纳的回答

Matt J 2017-7-27

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编辑：Matt J 2017-7-27

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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))

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Matt J 2017-7-27

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 2017-7-28