How to get rid of Nan when calculating mean
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Hi I am trying to get rid of the nan's on my mean calculation, if I keep them I wont be able to calculate z-scores. I was using m1=nanmean(x) but when I look at the output I still have the nan's on the column.
Thanks in advance.
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Shubham Gupta
2019년 10월 25일
편집: Shubham Gupta
2019년 10월 25일
To find indeces with NaN values you can use:
ind = find(isnan(A)); % where A is the array
To delete NaN values from that array you can use:
A(ind) = []; % replacing A(ind) with empty array
I am not sure if that's what you want? But it would be helpful to share some data, so we can understand the problem more clearly and be able to solve
Fabio Freschi
2019년 10월 25일
Are you saying that the built-in function in failing in the calculation? It seems weird...
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Steven Lord
2019년 10월 25일
The 'omitnan' option will ignore NaN values in your data when computing the mean. If your data contains values that result in a NaN being computed during the process of computing the mean then you'll receive NaN.
x = [1 NaN 2 3];
meanNoNaN = mean(x, 'omitnan') % No NaN
x2 = [1 NaN Inf -Inf 2 3];
meanNaN = mean(x2, 'omitnan') % Is NaN despite 'omitnan'
The NaN in the second element of x2 is ignored in the mean calculation. So we're taking the mean of five values: 1, Inf, -Inf, 2, and 3. As part of that mean calculation we need to add those five elements together using sum (as is normal for the standard arithmetic mean) and adding Inf and -Inf together results in NaN.
It's a bit subtle, but note that the documentation for the 'omitnan' option in the documentation for the mean function states "Ignore all NaN values in the input." [emphasis added] not "You can never get a NaN from this function if you specify this option." or something to that effect.
If you need to avoid even computed NaN values you probably want to remove non-finite values from your data before computing the mean. Use isfinite to identify those non-finite values or use rmoutliers to eliminate them.
Fabio Freschi
2019년 10월 25일
Following Steven Lord answer I would usE
mean(A(isfinite(A)))
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