df.groupby(by=["b"], dropna=False).sum() returns"groupby() got an unexpected keyword argument 'dropna'"

[vahanm01]

Using the official documentation for groupby: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.groupby.html

the dropna=False argument does not work even with given example.

l = [[1, 2, 3], [1, None, 4], [2, 1, 3], [1, 2, 2]]
df = pd.DataFrame(l, columns=["a", "b", "c"])
df.groupby(by=["b"], dropna=False).sum()

[arw2019]

This runs on master:

In [2]: l = [[1, 2, 3], [1, None, 4], [2, 1, 3], [1, 2, 2]] 
   ...: df = pd.DataFrame(l, columns=["a", "b", "c"]) 
   ...: df.groupby(by=["b"], dropna=False).sum()                                                                                                                                                                   
Out[2]: 
     a  c
b        
1.0  2  3
2.0  2  5
NaN  1  4

It was fixed in #35078

[rhshadrach]

Thanks for reporting this, the fix will be available when 1.2 is released.

[janalysis]

Any workarounds while I wait for 1.2?

[arw2019]

in case that helps 1.2 is out within a few weeks (and actually the release candidate is out already if you feel like trying it out)

[rhshadrach]

For a workaround, can replace nan with a value that doesn't exist in the frame, do the groupby, and then refill will nan.

[pramodh941]

The issue still persists.
"TypeError: group() got an unexpected keyword argument 'dropna'"