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

Asked By: Keiku

||

Source

Answer 1

The following will accomplish this:

df["elapsed_months"] = ((df["today"] - df["date"]).
                        map(lambda x: round(x.days/30)))


# Out[34]: 
#         date      today  elapsed_months
# 0 2016-10-11 2016-12-02               2
# 1 2016-11-01 2016-12-02               1

Answered By: J Darbyshire

Answer 2

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

Answered By: Psidom

Answer 3

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

Answered By: Michael Stokes

Answer 4

In a simpler way, it can also be calculated using the to_period function in pandas.

pd.to_datetime('today').to_period('M') - pd.to_datetime('2020-01-01').to_period('M')
# [Out]:
# <7 * MonthEnds>

In case, you just want the integer value just use (<above_code>).n

Answered By: aks

Answer 5

Update for pandas 1.3

If you want integers instead of MonthEnd objects:

df['elapsed_months'] = df.today.dt.to_period('M').view(dtype='int64') - df.date.dt.to_period('M').view(dtype='int64')
df

# Out[11]: 
#         date      today  elapsed_months
# 0 2016-10-11 2016-12-02               2
# 1 2016-11-01 2016-12-02               1

This works with pandas 1.1.1:

df['elapsed_months'] = df.today.dt.to_period('M').astype(int) - df.date.dt.to_period('M').astype(int)
df

# Out[11]: 
#         date      today  elapsed_months
# 0 2016-10-11 2016-12-02               2
# 1 2016-11-01 2016-12-02               1

Answered By: Pawel Kranzberg

Answer 6

Use can use .n to get the number of months as an integer:

(pd.to_datetime('today').to_period('M') - pd.to_datetime('2020-01-01').to_period('M')).n

On a dataframe, you can use it with .apply:

df["n_months"] = (df["date1"].dt.to_period("M") - df["date2"].dt.to_period("M")).apply(lambda x: x.n)

Also takes care of pandas 1.3.2 int conversion issue and any rounding issues with converting to ints earlier.

Answered By: firepod

Answer 7

If you don’t mind ignoring the days, you can use numpy functionality:

import numpy as np

df['elapsed month'] = (df.date.values.astype('datetime64[M]')-    
                    df.today.values.astype('datetime64[M]')) 
                          / np.timedelta64(1,'M')

Answered By: stedes

Pandas Timedelta in months

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