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

2018 4 20
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更新时间：2018 9 20
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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!

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Akira Agata
Akira Agata 2018-4-20

1 个投票

Another possible solution:

YourCellArray = cellfun(@replaceZeroWithNan,YourCellArray,'UniformOutput',false);

function D = replaceZeroWithNan(D)
idx = D == 0;
D(idx) = nan;
end

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Guillaume
Guillaume 2018-4-20

Yes, this is the solution I meant by: or cellfun, but in this case, you'd have to use a .m function not an anonymous function

Avishay Assayag
Avishay Assayag 2018-4-20
Thanks for your answer! That's an excellent solution for the general case. Kudos!
Guillaume
Guillaume 2018-4-20
编辑：Guillaume 2018-4-20

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

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Guillaume
Guillaume 2018-4-20
编辑：Guillaume 2018-4-20

0 个投票

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;

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Avishay Assayag
Avishay Assayag 2018-4-20
The first line of your code do nothing to my cell array.
Guillaume
Guillaume 2018-4-20

I'm not sure what you're referring to with "first line of code". The general solution does what you asked:

>> yourcellarray = {0:5, -1:1, eye(3)};
>> celldisp(yourcellarray)
yourcellarray{1} =
     0     1     2     3     4     5
yourcellarray{2} =
    -1     0     1
yourcellarray{3} =
     1     0     0
     0     1     0
     0     0     1

We have a cell array with 3 matrices of different size, with some 0s in them. Let's try the code:

>> for idx = 1:numel(yourcellarray)
   m = yourcellarray{idx};
   m(m == 0) = NaN;
   yourcellarray{idx} = m;
end
>> celldisp(yourcellarray)
yourcellarray{1} =
   NaN     1     2     3     4     5
yourcellarray{2} =
    -1   NaN     1
yourcellarray{3} =
     1   NaN   NaN
   NaN     1   NaN
   NaN   NaN     1

All the zeros have been changed to NaN.

Avishay Assayag
Avishay Assayag 2018-4-20
Sorry for being unclear. I meant to the example with m = cat(4, yourcellarray) which didn't work for me.
Guillaume
Guillaume 2018-4-20
Thanks, Stephen. Silly but crucial typo, now fixed in the original post.

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