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mean value of a group of data with NaNs

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Leon
Leon on 27 Jun 2016
Commented: the cyclist on 28 Jun 2016
I have a matrix A with the dimension of 12x360x180. It stores 12 values at each grid of the 360x180. Now I want to calculate the average value at each grid point. The problem is that there are unknown number of NaNs. Sometimes all 12 values are non-NaNs, sometimes all of them are NaNs, sometimes a portion of the 12 values are NaNs.
In this case, how do I estimate the mean values at each point of the grid? I know if there are no NaNs, the calculation is as easy as B = mean(A);
Thank you.

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Accepted Answer

the cyclist
the cyclist on 27 Jun 2016
Edited: the cyclist on 27 Jun 2016
If you have the Statistics and Machine Learning Toolbox, you can use the nanmean function, which computes the mean while ignoring NaNs.

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Leon
Leon on 27 Jun 2016
Many thanks! It works very well!
I happen to have Statistics toolbox on my version of the Matlab :-)

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More Answers (2)

Chris Turnes
Chris Turnes on 27 Jun 2016
You can also just use the 'omitnan' option in "mean":
A = [1 0 0 1 NaN 1 NaN 0];
M = mean(A,'omitnan')
M =
0.5000

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Leon
Leon on 27 Jun 2016
Wow, this is refreshing! Thank you for sharing!
Chris Turnes
Chris Turnes on 27 Jun 2016
By the way, other related functions support this input argument as well: sum, var, std, median, and max and min (max and min have 'omitnan' as the default option). The mov* functions (if you have R2016a or newer) also support this argument.
the cyclist
the cyclist on 28 Jun 2016
I was unaware of this option! Very nice. Please accept my upvote.

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the cyclist
the cyclist on 27 Jun 2016
If you do not have that Statistics and Machine Learning Toolbox, this should work:
notNan = not(isnan(A));
B = zeros(size(A));
B(notNan) = A(notNan);
sum(B)./sum(notNan)
You can do the sum over different dimensions, as required.

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Leon
Leon on 27 Jun 2016
Brilliant! Good to have this alternative method for folks who do not have the Statistic toolbox and come to this page through search engine.

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