BUG: Constructing DataFrame from list of series with MultiIndex, if MultiIndex contains (NaN, NaN)

[sziem]

• I have checked that this issue has not already been reported.

• I have confirmed this bug exists on the latest version of pandas.

• (optional) I have confirmed this bug exists on the master branch of pandas.

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I have observed a few issues, when constructing a dataframe from a list of series with multiindices, if any multiindex contains (NaN, NaN).

Code Sample, a copy-pastable example

1. In this case I would expect the columns of the dataframe to be the union of the indices of series1 and series2, but ("a", "b") is missing.

import pandas as pd
series1 = pd.Series((1,), index=pd.MultiIndex.from_tuples(((None, None),)))
series2 = pd.Series((10, 20), index=pd.MultiIndex.from_tuples(((None, None), ("a", "b"))))
df = pd.DataFrame([series1, series2])
#   NaN
#   NaN
# 0   1
# 1  10

2. There is another issue, when series1 and series2 are flipped. The columns appear to be correct, but in this case I would expect the value NaN at index 1, column (a,b)):

df = pd.DataFrame([series2, series1])
#     a NaN
#     b NaN
# 0  20  10
# 1   1   1

Problem description

Expected Output

df = pd.DataFrame([series1, series2])
#     a NaN
#     b NaN
# 0  NaN 1
# 1  20  10

and the reverse case

df = pd.DataFrame([series2, series1])
#     a NaN
#     b NaN
# 0  20  10
# 1  NaN 1

This behavior is not limited to both series containing the MultiIndex (NaN, NaN). Here is an example, where only one series is constructed with (None, None). The column ("a", "b") is missing.

import pandas as pd
series1 = pd.Series((1,), index=pd.MultiIndex.from_tuples((("a", None),)))
series2 = pd.Series((10, 20), index=pd.MultiIndex.from_tuples(((None, None), ("a", "b"))))
df = pd.DataFrame([series1, series2])
#      a   NaN
#    NaN   NaN
# 0  1.0   NaN
# 1  NaN  10.0

Output of pd.show_versions()

INSTALLED VERSIONS

commit : f346574
python : 3.9.6.final.0
python-bits : 64
OS : Linux
OS-release : 5.10.59-1-MANJARO
Version : #1 SMP PREEMPT Sun Aug 15 13:11:32 UTC 2021
machine : x86_64
processor :
byteorder : little
LC_ALL : None
LANG : en_US.utf8
LOCALE : en_US.UTF-8

pandas : 0+untagged.1.gf346574
numpy : 1.21.1
pytz : 2021.1
dateutil : 2.8.2
pip : 21.1.3
setuptools : 57.4.0
Cython : None
pytest : None
hypothesis : None
sphinx : None
blosc : None
feather : None
xlsxwriter : None
lxml.etree : None
html5lib : None
pymysql : None
psycopg2 : None
jinja2 : None
IPython : 7.26.0
pandas_datareader: None
bs4 : None
bottleneck : None
fsspec : None
fastparquet : None
gcsfs : None
matplotlib : 3.4.3
numexpr : None
odfpy : None
openpyxl : None
pandas_gbq : None
pyarrow : None
pyxlsb : None
s3fs : None
scipy : None
sqlalchemy : None
tables : None
tabulate : None
xarray : None
xlrd : None
xlwt : None
numba : None

[simonjayhawkins]

Thanks @sziem for the report.

There appears to be a change of behaviour here since 1.2.5 although the results on 1.2.5 look iffy.

>>> df = pd.DataFrame([series1, series2])
/home/simon/miniconda3/envs/pandas-1.2.5/lib/python3.9/site-packages/pandas/core/indexes/multi.py:3587: RuntimeWarning: The values in the array are unorderable. Pass `sort=False` to suppress this warning.
  uniq_tuples = lib.fast_unique_multiple([self._values, other._values], sort=sort)
>>> print(df)
  NaN       a
  NaN NaN   b
0   1   1   1
1  10  10  20
>>> 

However the last example does appear to give the correct result with 1.2.5?

>>> df = pd.DataFrame([series1, series2])
/home/simon/miniconda3/envs/pandas-1.2.5/lib/python3.9/site-packages/pandas/core/indexes/multi.py:3587: RuntimeWarning: The values in the array are unorderable. Pass `sort=False` to suppress this warning.
  uniq_tuples = lib.fast_unique_multiple([self._values, other._values], sort=sort)
>>> print(df)
     a   NaN     a
   NaN   NaN     b
0  1.0   NaN   1.0
1  NaN  10.0  20.0
>>> 

will label as a regression for the last case pending further investigation.

[simonjayhawkins]

However the last example does appear to give the correct result with 1.2.5?

maybe not. df.loc[0, ("a", "b")] should be NaN I think.

[phofl]

This is related to #37222

We changed the code paths we are running through for 1.3

[simonjayhawkins]

changing milestone to 1.3.5