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Find and replace values in a cell array containing 3-D matrixes with numeric values

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Hi, I have a cell array with 6 cells, each cell containing a 3-D matrix of 640x480x30. I'm trying to find all zeros (0) in the cell array and replace them with NaNs.
In the general case, I have a cell array (MyCellArray) with -K- cells, each cell containing an arbitrary vector or matrix in an arbitrary size, with numeric values. I'm trying to find all places where MyCellArray contains the value of X (some number) and replace it with Y (some number).
Is there a way to do it without a for loop? Something like (that of course can't work) MyCellArray{MyCellArray==0}=NaN
Thanks!

Réponse acceptée

Akira Agata
Akira Agata le 20 Avr 2018

Another possible solution:

YourCellArray = cellfun(@replaceZeroWithNan,YourCellArray,'UniformOutput',false);
function D = replaceZeroWithNan(D)
idx = D == 0;
D(idx) = nan;
end
  3 commentaires
Avishay Assayag
Avishay Assayag le 20 Avr 2018
Thanks for your answer! That's an excellent solution for the general case. Kudos!
Guillaume
Guillaume le 20 Avr 2018
Modifié(e) : Guillaume le 20 Avr 2018

Note that, while more concise, cellfun is generally slower than an explicit loop.

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Plus de réponses (1)

Guillaume
Guillaume le 20 Avr 2018
Modifié(e) : Guillaume le 20 Avr 2018

In the generic case, it is not possible to do it without a loop (or cellfun, but in this case, you'd have to use a .m function not an anonymous function:

for idx = 1:numel(yourcellarray)
   m = yourcellarray{idx};
   m(m == 0) = NaN;
   yourcellarray{idx} = m;
end

In your example case, where all the matrices are the same size, then you'd be better off not using a cell array but a 4-D matrix. The replacement is then trivial:

m = cat(4, yourcellarray{:});
m(m == 0) = NaN;
  5 commentaires
Guillaume
Guillaume le 20 Avr 2018
Thanks, Stephen. Silly but crucial typo, now fixed in the original post.

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