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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!

Respuesta aceptada

Akira Agata
Akira Agata el 20 de Abr. de 2018

Another possible solution:

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

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

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Más respuestas (1)

Guillaume
Guillaume el 20 de Abr. de 2018
Editada: Guillaume el 20 de Abr. de 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 comentarios
Guillaume
Guillaume el 20 de Abr. de 2018
Thanks, Stephen. Silly but crucial typo, now fixed in the original post.

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