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How can I average multiple matrices by element, ignoring NaN values, to create a new matrix of the same size?

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Patrick
Patrick on 16 Oct 2013
Commented: Jerry Gregoire on 14 Apr 2015
I have 248 matrices, each of size 72x144. There are some NaN values inserted randomly here and there in these matrices. I would like to create a new matrix, also of size 72x144, which is an average by element of the originals, but ignoring the NaN values. For example:
Original:
matrixA = [2 3 5 | 3 NaN 1 | 2 4 3]
matrixB = [3 4 NaN | 1 2 5 | NaN 3 5]
matrixC = [NaN 2 3 | 2 5 3 | 1 2 1]
Want:
matrixAvg = [2.5 3 4 | 2 3.5 3 | 1.5 3 3]
Is there a simple way for me to do this?

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Patrick
Patrick on 16 Oct 2013
I'm not sure what that means exactly, but the matrices are in my workspace, entitled "tsjul100," "tsjul103," "tsjul106"..."tsjul121," "tsjul200," etc., representing a 72x144 set of data taken every 3 hours for 31 days. Does that answer your question?

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Accepted Answer

Yannick
Yannick on 16 Oct 2013
Hi, if you have Statistics Toolbox, you can use nanmean as follows:
matrixA = [2 3 5 ; 3 NaN 1 ; 2 4 3];
matrixB = [3 4 NaN ; 1 2 5 ; NaN 3 5];
matrixC = [NaN 2 3 ; 2 5 3 ; 1 2 1];
allData = cat(3,matrixA,matrixB,matrixC);
nanmean(allData,3);

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Jerry Gregoire
Jerry Gregoire on 14 Apr 2015
You can also use mean with 'omitnan'
S = mean(..., MISSING) specifies how NaN (Not-A-Number) values are
treated. The default is 'includenan':
'includenan' - the mean of a vector containing NaN values is also NaN.
'omitnan' - the mean of a vector containing NaN values is the mean
of all its non-NaN elements. If all elements are NaN,
the result is NaN.

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More Answers (1)

Azzi Abdelmalek
Azzi Abdelmalek on 16 Oct 2013
Edited: Azzi Abdelmalek on 16 Oct 2013
Edit
matrixA = [2 3 5 ; 3 NaN 1 ; 2 4 3]
matrixB = [3 4 NaN ; 1 2 5 ; NaN 3 5]
matrixC = [NaN 2 3 ; 2 5 3 ; 1 2 1]
A={matrixA matrixB matrixC}
B=cat(3,A{:})
idx=find(isnan(B))
C=ones(size(B));
C(idx)=0;
B(idx)=0;
out=sum(B,3)./sum(C,3)

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Azzi Abdelmalek
Azzi Abdelmalek on 16 Oct 2013
The code replace NaN by zeros, but when calculating the mean, it is ignored, for example
[ nan 2 3] % is replaced by
[0 2 3]
% to calculate the mean
sum([0 2 3])./sum([0 1 1]) % =5/2=2.5

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