Using datetime.timedelta to add years

Question

I am doing some time calculations in Python.

Goal:

Part of this is trying to :

Given a date, add time interval (X years, X months, X weeks), return date

ie

• input args: input_time (datetime.date), interval (datetime.timedelta)
• return: datetime.date

I looked at the datetime and datetime.timedelta docs

class datetime.timedelta(days=0, seconds=0, microseconds=0, milliseconds=0, minutes=0, hours=0, weeks=0)¶.

These seem to work well if I want to add a certain number of hours or weeks. However,

Problem:

• I am trying to implement an operation such as date + 1 year and can't figure it out

E.g.

start = datetime.datetime(2000, 1, 1)
# expected output: datetime.datetime(2001, 1, 1)


# with the weeks, etc arguments given in timedelta, this fails unsurprisingly e.g 
start + datetime.timedelta(weeks = 52)

# returns datetime.datetime(2000, 12, 30, 0, 0)

Question

• Is this year-based operation possible with the basic tools of datetime - if so, how would I go about it?

• I realize that for the year example, I could just do start.replace(year = 2001), but that approach will fail if I have months or weeks as input.

• From my understanding, the dateutil library has more advanced features, but I was not clear how well it interacts with the in-built datetime objects.

I have reviewed this similar question but it did not help me with this.

Any help is much appreciated!

Running Python 3.6.5 on MacOs.

Answer 1

timedelta does not support years, because the duration of a year depends on which year (for example, leap years have Feb 29).

You could use a relativedelta instead, which does support years and takes into account the baseline date for additions:

>>> from dateutil.relativedelta import relativedelta
>>> now = datetime.now()
>>> now 
datetime.datetime(2019, 1, 27, 19, 4, 11, 628081)
>>> now + relativedelta(years=1)
datetime.datetime(2020, 1, 27, 19, 4, 11, 628081)