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

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?

[Azzi Abdelmalek]

How your matrices are stored?

[Patrick]

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?

[Vivek Selvam]

Does your matrix have any element as zero?

[Patrick]

There are no elements as zero, only NaN.

Answer 1

[Yannick]

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);

[Patrick]

Thank you so much, that worked immediately!

[Jerry Gregoire]

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.

Answer 2

[Azzi Abdelmalek]

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)

[Patrick]

This is giving me a new matrix of size 72x144x248. I need one of only 72x144.

[Azzi Abdelmalek]

No, you can check for the above example,

size(out)

[Azzi Abdelmalek]

There is just one mistake

out=sum(B,3)./sum(C,3)  % instead of .*

 out =

    2.5000    3.0000    4.0000
    2.0000    3.5000    3.0000
    1.5000    3.0000    3.0000

[Patrick]

Actually, yes, you are correct. I was looking at "C" as my final result instead of "out." However, now I have noticed that "out" is the correct size, but contains the wrong numbers. To try and solve this problem, I decided:

outfinal=out./248

This is giving me approximately correct numbers, but with some anomalously low numbers as well. Can you explain what I might be doing wrong?

[Azzi Abdelmalek]

Look at edited answer, I've mistaken by using .* instead of ./

[Patrick]

Has your code replaced NaN values with zeroes?

[Azzi Abdelmalek]

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

[Patrick]

Thank you for your help Azzi, I have solved the problem.