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The current behaviour returns the sum of default timestamp with the number of unique count of dates in each year. The expected output should have been the number of unique counts for the date in each year which is consistent with the behavior of other Groupby.transform functions.
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Code Sample, a copy-pastable example
Problem description
The current behaviour returns the sum of default timestamp with the number of unique count of dates in each year. The expected output should have been the number of unique counts for the date in each year which is consistent with the behavior of other Groupby.transform functions.
Expected Output
Output of
pd.show_versions()
INSTALLED VERSIONS
commit : None
python : 3.7.7.final.0
python-bits : 64
OS : Linux
OS-release : 4.14.181-140.257.amzn2.x86_64
machine : x86_64
processor : x86_64
byteorder : little
LC_ALL : None
LANG : en_US.UTF-8
LOCALE : en_US.UTF-8
pandas : 1.0.4
numpy : 1.18.5
pytz : 2020.1
dateutil : 2.8.1
pip : 20.0.2
setuptools : 47.1.1.post20200604
Cython : None
pytest : None
hypothesis : None
sphinx : None
blosc : None
feather : None
xlsxwriter : None
lxml.etree : None
html5lib : None
pymysql : None
psycopg2 : None
jinja2 : 2.11.2
IPython : 7.15.0
pandas_datareader: None
bs4 : None
bottleneck : None
fastparquet : None
gcsfs : None
lxml.etree : None
matplotlib : 3.2.1
numexpr : 2.7.1
odfpy : None
openpyxl : 3.0.4
pandas_gbq : None
pyarrow : 0.17.1
pytables : None
pytest : None
pyxlsb : None
s3fs : None
scipy : 1.4.1
sqlalchemy : None
tables : None
tabulate : 0.8.7
xarray : None
xlrd : None
xlwt : None
xlsxwriter : None
numba : None
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