To find the average when there is NaN

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Thishan Dharshana Karandana Gamalathge
Kommentiert: Walter Roberson am 18 Aug. 2017
I want to find the average of each pairs. ie. avg of 2 &4, avg of NaN and 5. Answer should be 3 and 5. For the second pair, code should eliminate NaN, so avg should be 5. That's where I have the problem. Following is the code I wrote. Thanks.
A=[2 4 NaN 5];
j=1;
for i=1:2:4
B=A(i:i+1);
C=~isnan(B);
test(j)=mean(A(C));
i=i+1;
j=j+1;
end

Antworten (2)

Chad Greene
Chad Greene am 17 Aug. 2017
Bearbeitet: Chad Greene am 17 Aug. 2017
Hi Thishan,
On newer versions of Matlab, use
test(j)=mean(A(C),'omitnan');
If you have an older version of Matlab and you have the Statistics Toolbox, use
test(j)=nanmean(A(C));
Alternatively, if neither of those options work for you,
tmp = A(C);
test(j)= = mean(tmp(isfinite(tmp)));
  7 Kommentare
Image Analyst
Image Analyst am 17 Aug. 2017
Sure it makes sense. You can move along the vector an element at a time taking the means of pairs, no matter whether there are an odd or even number of elements. Just look:
m = [1,2,3,4,5]
pairMeans = conv(m, [.5,.5], 'valid')
m =
1 2 3 4 5
pairMeans =
1.5 2.5 3.5 4.5
(Note the above does not handle nans in the desired way).
Walter Roberson
Walter Roberson am 18 Aug. 2017
However, the user's for loop increments by 2, so that is not the pairs they meant. They also indicated they expected exactly 2 results, not 3.

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Image Analyst
Image Analyst am 17 Aug. 2017
Bearbeitet: Image Analyst am 17 Aug. 2017
This will work for any length of A, even odd numbers:
A=[2 4 NaN 5 6 8 9]; % 7 elements.
kernel = [1,1];
sumA = conv(A, kernel, 'same')
nonNaNA = ~isnan(A)
denom = conv(nonNaNA, kernel, 'same')
output = sumA ./ denom
output(isnan(output)) = [] % Remove nans.

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