Pandas Timedelta in months

Question

How can I calculate the elapsed months using pandas? I have write the following, but this code is not elegant. Could you tell me a better way?

import pandas as pd  df = pd.DataFrame([pd.Timestamp('20161011'),                    pd.Timestamp('20161101') ], columns=['date']) df['today'] = pd.Timestamp('20161202')  df = df.assign(     elapsed_months=(12 *                     (df["today"].map(lambda x: x.year) -                      df["date"].map(lambda x: x.year)) +                     (df["today"].map(lambda x: x.month) -                      df["date"].map(lambda x: x.month)))) # Out[34]:  #         date      today  elapsed_months # 0 2016-10-11 2016-12-02               2 # 1 2016-11-01 2016-12-02               1 

Answer 1

Update for pandas 0.24.0:

Since 0.24.0 has changed the api to return MonthEnd object from period subtraction, you could do some manual calculation as follows to get the whole month difference:

12 * (df.today.dt.year - df.date.dt.year) + (df.today.dt.month - df.date.dt.month)  # 0    2 # 1    1 # dtype: int64 

Wrap in a function:

def month_diff(a, b):     return 12 * (a.dt.year - b.dt.year) + (a.dt.month - b.dt.month)  month_diff(df.today, df.date) # 0    2 # 1    1 # dtype: int64 

---

Prior to pandas 0.24.0. You can round the date to Month with to_period() and then subtract the result:

df['elapased_months'] = df.today.dt.to_period('M') - df.date.dt.to_period('M')  df #         date       today  elapased_months #0  2016-10-11  2016-12-02                2 #1  2016-11-01  2016-12-02                1 

Answer 2

you could also try:

df['months'] = (df['today'] - df['date']) / np.timedelta64(1, 'M') df #      date      today    months #0 2016-10-11 2016-12-02  1.708454 #1 2016-11-01 2016-12-02  1.018501