How to segment one row matrix into multiple row matrix

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Lukas Poviser 2021-4-10

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https://ww2.mathworks.cn/matlabcentral/answers/798277-how-to-segment-one-row-matrix-into-multiple-row-matrix

评论： David Fletcher 2021-4-10

Hi, I do have for example this matrix [ 3 6 4 5 6 NaN 8 9 10 NaN NaN NaN 13 14 5 6 0 1 NaN NaN 2 3 NaN NaN NaN 9 1 5 NaN 2 8 NaN] and i would like to have matrix where each row would be each segment of values.

I mean like:

3 6 4 5 6

8 9 10

13 14 5 6 0 1

2 3

9 1 5

2 8

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Answer 1

David Fletcher 2021-4-10

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Ultimately you can't have a matrix of the segments since they are potentially of different lengths - you could have a cell array of them though. Something like this:

A=[ 3 6 4 5 6 NaN 8 9 10 NaN NaN NaN 13 14 5 6 0 1 NaN NaN 2 3 NaN NaN NaN 9 1 5 NaN 2 8 NaN]

nanLoc=isnan(A);
partial=[];
vecCount=1;
iter=1;
vectors={};
while iter<=numel(A)
    if (nanLoc(iter)==1)
        if ~isempty(partial)
            vectors{vecCount} = partial;
            vecCount=vecCount+1
        end
        partial=[];
    else
        partial=[partial A(iter)];
    end
iter=iter+1;
    
end

Running it gives

vectors =

1×6 cell array

{[3 6 4 5 6]} {[8 9 10]} {[13 14 5 6 0 1]} {[2 3]} {[9 1 5]} {[2 8]}

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Lukas Poviser 2021-4-10

And would it be possible to interpolate vectors? So every vector would have 100 values. And final size of matrice would be M(number of segments) x 100.

David Fletcher 2021-4-10

If you wished to pad them to the same length and store them in a matrix, then yes you could do that. However, you (presumably) could not know beforehand what size the segments would need to be padded to - saying that some trivial analysis of the nanLoc vector could give you such a value. Also whatever you used to pad the vectors couldn't be a value that would naturally occur in the vector (or you would have no way of knowing whether a pad value was genuine or naturally occuring). One final thing, as it often the case I have just noticed a problem in the code I've provided - if the original array doesn't end in a NaN, the final partial value will fail to be written to vectors cell array - but then I guess it does leave something for you to do...

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