CODEWITHSUNDEEP
pythonpandasdatetime
Pdubbs
Pdubbs

Reputation: 1987

How to add a year to a column of dates in pandas

I am attempting to add a year to a column of dates in a pandas dataframe, but when I use pd.to_timedelta I get additional hours & minutes. I know I could take the updated time and truncate the hours, but I feel like there must be a way to add a year precisely. My attempt as follows:

import pandas as pd
dates = pd.DataFrame({'date':['20170101','20170102','20170103']})
dates['date'] = pd.to_datetime(dates['date'], format='%Y%m%d')
dates['date2'] = dates['date'] +  pd.to_timedelta(1, unit='y')
dates

yields:

Out[1]: 
    date        date2
0   2017-01-01  2018-01-01 05:49:12
1   2017-01-02  2018-01-02 05:49:12
2   2017-01-03  2018-01-03 05:49:12

How can I add a year without adding 05:49:12 HH:mm:ss?

Upvotes: 23

Views: 40597

Answers (5)

Mostafa Elhassan Elsawy
Mostafa Elhassan Elsawy

Reputation: 1

You can access the components of a date (year, month and day) using code of the form dataframe["column"].dt.component.

For example, the month component is dataframe["column"].dt.month, and the year component is dataframe["column"].dt.year.

Upvotes: 0

ZaxR
ZaxR

Reputation: 5165

Edit: This works if you don't care about leap years, etc. Otherwise see jp_data_analysis's answer.

You can use 365 and unit='d':

pd.to_timedelta(365, unit='d')

Upvotes: 0

Omni
Omni

Reputation: 1022

Or convert datetime to date

dates['date'] = dates['date'].apply(lambda a: a.date())

Upvotes: 2

MaxU - stand with Ukraine
MaxU - stand with Ukraine

Reputation: 210972

In [99]: dates['date'] + pd.offsets.DateOffset(years=1)
Out[99]:
0   2018-01-01
1   2018-01-02
2   2018-01-03
Name: date, dtype: datetime64[ns]

leap year check:

In [100]: pd.to_datetime(['2011-02-28', '2012-02-29']) + pd.offsets.DateOffset(years=1)
Out[100]: DatetimeIndex(['2012-02-28', '2013-02-28'], dtype='datetime64[ns]', freq=None)

Upvotes: 41

jpp
jpp

Reputation: 164823

You can normalize via pd.Series.dt.normalize:

dates['date2'] = (dates['date'] +  pd.to_timedelta(1, unit='y')).dt.normalize()

Upvotes: 4

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