Extracting just Month and Year separately from Pandas Datetime column

Question

I have a Dataframe, df, with the following column:

df[\'ArrivalDate\'] =
...
936   2012-12-31
938   2012-12-29
965   2012-12-31
966   2012-12-31
967   2012-1         


        

Answer 1

You can directly access the year and month attributes, or request a datetime.datetime:

In [15]: t = pandas.tslib.Timestamp.now()

In [16]: t
Out[16]: Timestamp('2014-08-05 14:49:39.643701', tz=None)

In [17]: t.to_pydatetime() #datetime method is deprecated
Out[17]: datetime.datetime(2014, 8, 5, 14, 49, 39, 643701)

In [18]: t.day
Out[18]: 5

In [19]: t.month
Out[19]: 8

In [20]: t.year
Out[20]: 2014

One way to combine year and month is to make an integer encoding them, such as: 201408 for August, 2014. Along a whole column, you could do this as:

df['YearMonth'] = df['ArrivalDate'].map(lambda x: 100*x.year + x.month)

or many variants thereof.

I'm not a big fan of doing this, though, since it makes date alignment and arithmetic painful later and especially painful for others who come upon your code or data without this same convention. A better way is to choose a day-of-month convention, such as final non-US-holiday weekday, or first day, etc., and leave the data in a date/time format with the chosen date convention.

The calendar module is useful for obtaining the number value of certain days such as the final weekday. Then you could do something like:

import calendar
import datetime
df['AdjustedDateToEndOfMonth'] = df['ArrivalDate'].map(
    lambda x: datetime.datetime(
        x.year,
        x.month,
        max(calendar.monthcalendar(x.year, x.month)[-1][:5])
    )
)

If you happen to be looking for a way to solve the simpler problem of just formatting the datetime column into some stringified representation, for that you can just make use of the strftime function from the datetime.datetime class, like this:

In [5]: df
Out[5]: 
            date_time
0 2014-10-17 22:00:03

In [6]: df.date_time
Out[6]: 
0   2014-10-17 22:00:03
Name: date_time, dtype: datetime64[ns]

In [7]: df.date_time.map(lambda x: x.strftime('%Y-%m-%d'))
Out[7]: 
0    2014-10-17
Name: date_time, dtype: object