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

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!

 채택된 답변

Another possible solution:

YourCellArray = cellfun(@replaceZeroWithNan,YourCellArray,'UniformOutput',false);
function D = replaceZeroWithNan(D)
idx = D == 0;
D(idx) = nan;
end

댓글 수: 3

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

Thanks for your answer! That's an excellent solution for the general case. Kudos!

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

댓글을 달려면 로그인하십시오.

추가 답변 (1개)

Guillaume
Guillaume 2018년 4월 20일
편집: Guillaume 2018년 4월 20일

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;

댓글 수: 5

The first line of your code do nothing to my cell array.

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.

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

댓글을 달려면 로그인하십시오.

카테고리

도움말 센터File Exchange에서 Logical에 대해 자세히 알아보기

질문:

2018년 4월 20일

편집:

2018년 9월 20일

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by