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  • nansum vs nanmean for all-nan vectors · 3 ✖

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id html_url issue_url node_id user created_at updated_at ▲ author_association body reactions performed_via_github_app issue
482162700 https://github.com/pydata/xarray/issues/2889#issuecomment-482162700 https://api.github.com/repos/pydata/xarray/issues/2889 MDEyOklzc3VlQ29tbWVudDQ4MjE2MjcwMA== fujiisoup 6815844 2019-04-11T15:28:58Z 2022-01-05T21:59:48Z MEMBER

Thanks @mathause I also think the current behavior is not perfect but the best.

I would expect both to return np.nan

I expect that np.nansum(ds) is equivalent to np.sum(not nan values) and thus should be 0, while np.mean should be NaN as @dcherian pointed out.

To me, the future average function would also return np.nan for all nan slices.

 
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   nansum vs nanmean for all-nan vectors 432074821
482178469 https://github.com/pydata/xarray/issues/2889#issuecomment-482178469 https://api.github.com/repos/pydata/xarray/issues/2889 MDEyOklzc3VlQ29tbWVudDQ4MjE3ODQ2OQ== mathause 10194086 2019-04-11T16:08:02Z 2019-04-11T16:08:02Z MEMBER

Thanks for the feedback. I think I see the reasoning behind it now.

 
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   nansum vs nanmean for all-nan vectors 432074821
482158883 https://github.com/pydata/xarray/issues/2889#issuecomment-482158883 https://api.github.com/repos/pydata/xarray/issues/2889 MDEyOklzc3VlQ29tbWVudDQ4MjE1ODg4Mw== dcherian 2448579 2019-04-11T15:19:42Z 2019-04-11T15:19:42Z MEMBER

This is correct though isn't it?

Mean = nansum(not nan values)/count(not nan values) = 0/0 

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   nansum vs nanmean for all-nan vectors 432074821

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