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nlargest gives a zero-row dataframe when ordering columns are all NaN #28984

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karldw opened this issue Oct 15, 2019 · 1 comment · Fixed by #43060
Closed

nlargest gives a zero-row dataframe when ordering columns are all NaN #28984

karldw opened this issue Oct 15, 2019 · 1 comment · Fixed by #43060
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Apply Apply, Aggregate, Transform, Map Enhancement Groupby
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@karldw
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karldw commented Oct 15, 2019

Code Sample, a copy-pastable example if possible

import pandas as pd
from numpy import nan
df = pd.DataFrame({'grp': [1, 1, 2, 2], 'y': [1, 0, 2, 5], 'z': [1, 2, nan, nan]})
df.groupby('grp').apply(lambda grp_df: grp_df.nlargest(1, 'z'))
#        grp  y    z
# grp               
# 1   1    1  0  2.0
# 
# (Group 2 is gone!)

Problem description

When the values of the ordering variables are all missing, the nlargest and nsmallest methods return a zero-row dataframe. This behavior is particularly unexpected when applying over groups, since it silently omits groups with all-NaN values. I think it would be better to return the requested number of rows, with NaN as appropriate.

Put differently, this is a case where nlargest differs from the corresponding sort_values(...).head(...) code.

df.groupby('grp').apply(lambda x: x.sort_values('z', ascending=False).head(1))
#        grp  y    z
# grp               
# 1   1    1  0  2.0
# 2   2    2  2  NaN

(I'm aware that a better way to write that sort_values line would be to skip the apply and write df.sort_values('z', ascending=False).groupby('grp').head(1), which gives a similar result, but better index.)

I've talked about grouped dataframes because that seems like a more pernicious problem, but the behavior is the same with ungrouped dataframes.
The problem is also the same for nsmallest and nlargest, and it doesn't matter if there are multiple ordering columns, as long as they're all NaN.

Related, but not identical issues:
#23993 (requesting nlargest and nsmallest methods for grouped dataframes)
#21426 (bug with unsigned integers)
#12694 (NaN in Series.argsort)

Expected Output

df.groupby('grp').apply(lambda grp_df: grp_df.nlargest(1, 'z'))
#        grp  y    z
# grp               
# 1   1    1  0  2.0
# 2   2    2  2  NaN

Output of pd.show_versions()

INSTALLED VERSIONS

commit : None
python : 3.7.3.final.0
python-bits : 64
OS : Linux
OS-release : 5.0.0-31-generic
machine : x86_64
processor : x86_64
byteorder : little
LC_ALL : None
LANG : en_US.UTF-8
LOCALE : en_US.UTF-8

pandas : 0.25.1
numpy : 1.17.2
pytz : 2019.3
dateutil : 2.8.0
pip : 19.2.3
setuptools : 41.4.0
Cython : None
pytest : None
hypothesis : None
sphinx : None
blosc : None
feather : None
xlsxwriter : None
lxml.etree : None
html5lib : None
pymysql : None
psycopg2 : None
jinja2 : 2.10.3
IPython : 7.8.0
pandas_datareader: None
bs4 : None
bottleneck : None
fastparquet : None
gcsfs : None
lxml.etree : None
matplotlib : 2.2.4
numexpr : 2.7.0
odfpy : None
openpyxl : None
pandas_gbq : None
pyarrow : 0.15.0
pytables : None
s3fs : None
scipy : 1.3.1
sqlalchemy : None
tables : None
xarray : None
xlrd : None
xlwt : None
xlsxwriter : None

@h-vetinari
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I think the correct approach would be to add a dropna parameter (defaulting to True) that exists for several other methods already.

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