Pandas: filter for 2nd day of the month

October 5, 2022

I have the following data.

            value
date                
2010-10-01  100
2010-10-02  200
2010-10-03  300
2012-10-02  100
2012-10-03  200
2012-10-04  300
2022-10-03  500
2022-10-04  600

I would like to filter for the 2nd day of the month for each year. So i should end up with

date        value       
2010-10-02  200
2012-10-03  200
2022-10-04  600

Having a difficult time figuring out the most efficient way to do this without iteration/loops or lamdas. I thought using the following should work, but does not.

import pandas as pd
from pandas.tseries.offsets import BDay, BusinessDay, BMonthBegin
df = df[df.index == df.index + BMonthBegin(0) + BDay(1)]

Any help here would be appreciated.

>Solution :

Use DatetimeIndex.to_period for months periods and for second days use GroupBy.cumcount with compare second values in boolean indexing:

df = df[df.groupby(df.index.to_period('m')).cumcount().eq(1)]

print (df)
            value
date             
2010-10-02    200
2012-10-03    200
2022-10-04    600