Inconsistent behaviour with mean on masked array with and without dimensions given.

The mean gives different results depending on how it is taken on a multi-dimensional masked array. The following is a code-snippet that showcases it:

array = sc.DataArray(
    sc.array(dims=['x', 'y'], values=[[1.0, 1.0, 1.0, 1.0], [0.0, np.inf, 0.0, 1.0]]),
    masks={'mask' : sc.array(dims=['x', 'y'], values=[[False, False, False, False], [False, True, False, False]])}
    )
array.mean() # gives 0.714286, (5/7) as it should
array.mean(dim=['x', 'y']) # gives 0.75
array.mean(dim=['y', 'x'] # gives 0.66667

I guess this is because when we say .mean(['x', 'y']), we do first mean along 'x', and then mean along 'y'.
If you follow this reasoning, then you can understand where the results come from.

You get the same results from numpy:

import numpy as np

a = np.ma.masked_array(data=[[1.0, 1.0, 1.0, 1.0], [0.0, np.inf, 0.0, 1.0]],
                   mask=[[False, False, False, False], [False, True, False, False]])

a.mean()  # np.float64(0.7142857142857143)
a.mean(axis=0).mean(axis=0)  # np.float64(0.75)
a.mean(axis=1).mean(axis=0)  # np.float64(0.6666666666666666)

However, numpy differs with us when you pass both axes in one go:

a.mean(axis=(0, 1))  # np.float64(0.7142857142857143)
a.mean(axis=(1, 0))  # np.float64(0.7142857142857143)

which I guess is the desired behaviour?

So we should detect if the list of dims passed comprises all dims, and then just return the .mean() instead of doing the dims one by one?

So we should detect if the list of dims passed comprises all dims, and then just return the .mean() instead of doing the dims one by one?

That wouldn't be enough. E.g., given a 3d array and passing 2 dims, we would still differ from numpy.

That wouldn't be enough. E.g., given a 3d array and passing 2 dims, we would still differ from numpy.

And that is exactly my use-case. I'd like the mean to work like .mean() on each value of the third dimension. So something like:

for i in array.dims['t']:
   new_array['t', i] = array['t',i].mean()

Should the mean operation then just be implemented as the sum divided by volume, instead of doing mean on one dimension at a time?

Should the mean operation then just be implemented as the sum divided by volume, instead of doing mean on one dimension at a time?

I am not sure what exactly you mean by volume, but if volume is the number of elements that were summed up, then yeah, that's the behavior I would expect :)

Any updates on this issue?

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Inconsistent behaviour with `mean` on masked array with and without dimensions given. #3841

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Inconsistent behaviour with mean on masked array with and without dimensions given. #3841

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nvaytet commented on Feb 11, 2026

jl-wynen commented on Feb 12, 2026

damskii9992 commented on Feb 12, 2026

nvaytet commented on Feb 17, 2026

damskii9992 commented on Feb 17, 2026

damskii9992 commented on Jun 3, 2026

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Inconsistent behaviour with `mean` on masked array with and without dimensions given. #3841