Row vector, array matrix
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You can read this link. The code is good for the 1x72 double but it's not a general code because I could have 1xn double. How can I do in this situation? The question is that it's not always a 8x8 matrix but it's depend on the number of zero in the 1xn double. The problem is that I have in general 1xn double and not 1x72. So the reshape (8,8) it's not good in general. The important thing is that when there is a zero value, it's start a new rows: always. This is the important thing.
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Mischa Kim
2014-2-21
编辑：Mischa Kim
2014-2-21
Does the zero entry appear periodically? That is every nth element is zero?
The issue here is that if there is no periodicity you end up with rows of different length, which you cannot save in a matrix (without filling up the matrix with zeros). In that case you would have to use e.g. a cell array.
Francesco
2014-2-21
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Mischa Kim
2014-2-21
编辑：Mischa Kim
2014-2-21
OK. Provided that the zeros appear periodically just the way you have shown in your code, you could use
indz = find(A==0); % find positions of zeros in A
numr = indz(2)-indz(1)-1; % compute the constant(!) row lengths
A(A == 0) = [];
B = reshape(A,numr,length(A)/numr)
Again, if there is not a constant number of non-zero elements in each interval (that includes also the last interval) you end up with different row lengths. In that case you cannot build a matrix that satisfies your requirements.
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