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

Question

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?

score 38 · Accepted Answer · answered Feb 14 '18 at 21:51

38

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)

answered Feb 14 '18 at 21:51

MaxU - stand with Ukraine

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Good answer. Link to [docs](https://pandas.pydata.org/pandas-docs/stable/user_guide/timeseries.html#dateoffset-objects). Mind typos! That it is `years`, not `year`, or get a nasty out of bounds error. – Wtower Mar 01 '22 at 07:13

score 4 · Answer 2 · answered Feb 14 '18 at 21:50

4

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

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

answered Feb 14 '18 at 21:50

jpp

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This no longer works in new versions of pandas. – Jan Šimbera Dec 28 '21 at 18:16

score 2 · Answer 3 · answered Feb 14 '18 at 21:52

2

Or convert datetime to date

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

answered Feb 14 '18 at 21:52

Omni

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ZaxR · Answer 4 · 2018-02-14T21:58:30.300

0

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')

edited Feb 14 '18 at 21:58

answered Feb 14 '18 at 21:47

ZaxR

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Unfortunately I have a large number of dates and need to account for leap years. I'd like to preserve calendar date to account for holidays (New Year's day, Christmas, etc.) – Pdubbs Feb 14 '18 at 21:51
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Gotcha. jp_data_analysis's answer should work [and be fast] then. – ZaxR Feb 14 '18 at 21:54

score 0 · Answer 5 · edited Jun 02 '21 at 16:12

0

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.

edited Jun 02 '21 at 16:12

develth

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answered Jun 02 '21 at 13:12

Mostafa Elhassan Elsawy

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Good to know, but not useful if you need to modify: `ValueError: modifications to a property of a datetimelike object are not supported. Change values on the original.` – Wtower Mar 01 '22 at 07:07

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

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