BUG: Wrong grouping of categoricals when observed=True

[topper-123]

Setup:

>>> s1 = pd.Categorical([np.nan, 'a', np.nan, 'a'], categories=['a', 'b'])
>>> s2 = pd.Series([1, 2, 3, 4])
>>> df = pd.DataFrame({'s1': s1, 's2': s2})

Comparing results with observed=False and observed=True:

>>> df.groupby('s1').sum()  # ok
    s2
s1
a    6
b    0
>>> df.groupby('s1', observed=True).sum()
     s2
s1
NaN   6  # should not be shown
a     0  # should be 6

Notice the value are assigned wrongly.

Also, notice that NaN is now a possible label.

If the first value is not a Nan, the assignment works fine (but Nan is still a possible label)

>>> df[1:].groupby('s1', observed=True).sum()
     s2
s1
a     6
NaN   0

Problem description

The problem concerns when there are unobserved labels and the first value is Nan. If there are no unobserved values, everyting seems alright from my checks.

Nan should probably not be a possible label.

Expected Output

I would assume same output as when òbserved=False, but without unobserved labels:

>>> df.groupby('s1', observed=True).sum()
    s2
s1
a    6

Output of pd.show_versions()

INSTALLED VERSIONS

commit: None
python: 3.6.3.final.0
python-bits: 32
OS: Windows
OS-release: 10
machine: AMD64
processor: Intel64 Family 6 Model 78 Stepping 3, GenuineIntel
byteorder: little
LC_ALL: None
LANG: None
LOCALE: None.None

pandas: 0.23.0
pytest: 3.4.0
pip: 10.0.1
setuptools: 38.4.1
Cython: 0.26.1
numpy: 1.14.0
scipy: 1.0.0
pyarrow: None
xarray: None
IPython: 6.3.1
sphinx: 1.7.4
patsy: None
dateutil: 2.6.1
pytz: 2017.3
blosc: None
bottleneck: None
tables: 3.4.2
numexpr: 2.6.2
feather: None
matplotlib: 2.1.0
openpyxl: 2.4.8
xlrd: None
xlwt: None
xlsxwriter: None
lxml: 4.1.1
bs4: 4.6.0
html5lib: 1.0.1
sqlalchemy: None
pymysql: None
psycopg2: None
jinja2: 2.10
s3fs: None
fastparquet: None
pandas_gbq: None
pandas_datareader: None

[diegogarcilazo]

it seems to be related to issue #21133

[cwkwong]

I've also tested the above groupby() followed by first() and it seems to produce some unexpected results.

Setup:

>>> s1 = pd.Categorical([np.nan, 'a', np.nan, 'a'], categories=['a', 'b'])
>>> s2 = pd.Series([1,2,3,4])
>>> 
>>> df = pd.DataFrame({'s1':s1, 's2':s2})
>>> df
    s1  s2
0  NaN   1
1    a   2
2  NaN   3
3    a   4

Comparing results with observed=False and observed=True:

>>> df.groupby('s1').first()
     s2
s1     
a   2.0
b   NaN

>>> df.groupby('s1', observed=True).first()
      s2
s1      
NaN  2.0
a    NaN

When observed=True, the results is somewhat strange.

• Incorrect s2=NaN for s1=a. I think should be s2=2.0 ?
• NaN appearing in the groupby(...).first() result for s1? I would expect no NaN under s1?

Output of pd.show_versions()

INSTALLED VERSIONS

commit: None
python: 2.7.13.final.0
python-bits: 64
OS: Darwin
OS-release: 17.5.0
machine: x86_64
processor: i386
byteorder: little
LC_ALL: None
LANG: en_AU.UTF-8
LOCALE: None.None

pandas: 0.23.0
pytest: 3.3.1
pip: 9.0.1
setuptools: 18.2
Cython: 0.25.2
numpy: 1.14.0
scipy: 0.16.0
pyarrow: None
xarray: None
IPython: None
sphinx: None
patsy: None
dateutil: 2.7.3
pytz: 2014.4
blosc: None
bottleneck: 1.2.0
tables: 3.4.2
numexpr: 2.6.4
feather: None
matplotlib: 1.4.3
openpyxl: 1.8.6
xlrd: 0.9.3
xlwt: None
xlsxwriter: 0.9.6
lxml: 3.5.0
bs4: None
html5lib: None
sqlalchemy: 1.0.9
pymysql: None
psycopg2: 2.7.4 (dt dec pq3 ext lo64)
jinja2: 2.9.6
s3fs: None
fastparquet: None
pandas_gbq: None
pandas_datareader: None

[cgangwar11]

In [9]: import pandas as pd  
   ...: s1 = pd.Categorical([np.nan, 'a', np.nan, 'a'], categories=['a', 'b','c']) 
   ...: s2 = pd.Series([1,2,3,4]) 
   ...: df = pd.DataFrame({'s1':s1, 's2':s2})                                                                    

In [10]: df                                                                                                      
Out[10]: 
    s1  s2
0  NaN   1
1    a   2
2  NaN   3
3    a   4

In [11]: df.groupby('s1').first()                                                                                
Out[11]: 
     s2
s1     
a   2.0
b   NaN
c   NaN

In [12]: df.groupby('s1',observed=True).first()                                                                  
Out[12]: 
    s2
s1    
a    2

In [13]: