Nested mean function with omitnan
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Not sure if this behavior is intentional. When I have an array, let's say:
A=[1,NaN;3,4];
And I want to calculate the mean over the second dimension first, then the first one:
m=mean(A,[2,1],'omitnan');
The result is 2.66, which is clearly just (1+3+4)/3, and so it's the same as:
m=mean(A,'all','omitnan');
I'm aware that this is exactly what it says in the documentation, but it's still unexpected. What I would have expected would be the same as:
m=mean(mean(A,2,'omitnan'),1,'omitnan');
Which in this case gives the (imho correct) result of 2.25, so (1+(3+4)/2)/2
Am I missing something?
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Jan
2023-3-16
1 个投票
The problem is here: "What I would have expected would be the same as". This is not the way mean('omitnan') is working. So if you expect something else, you have to write the code to do it as you want it.
If it is explained in the documentation, there is no reason to expect something else. The behavior you expect would not match my intuition for [NaN, 1; NaN, 2].
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