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!
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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
Avishay Assayag
le 20 Avr 2018
Guillaume
le 20 Avr 2018
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
le 20 Avr 2018
Stephen23
le 20 Avr 2018
m = cat(4, yourcellarray{:});
Guillaume
le 20 Avr 2018
Thanks, Stephen. Silly but crucial typo, now fixed in the original post.
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