What's the simplest way to subtract a month from a date in Python?

Question

If only timedelta had a month argument in it\'s constructor. So what\'s the simplest way to do this?

EDIT: I wasn\'t thinking too hard about this as was poin

Answer 1

A variation on Duncan's answer (I don't have sufficient reputation to comment), which uses calendar.monthrange to dramatically simplify the computation of the last day of the month:

import calendar
def monthdelta(date, delta):
    m, y = (date.month+delta) % 12, date.year + ((date.month)+delta-1) // 12
    if not m: m = 12
    d = min(date.day, calendar.monthrange(y, m)[1])
    return date.replace(day=d,month=m, year=y)

Info on monthrange from Get Last Day of the Month in Python

Answer 2

You can use the third party dateutil module (PyPI entry here).

import datetime
import dateutil.relativedelta

d = datetime.datetime.strptime("2013-03-31", "%Y-%m-%d")
d2 = d - dateutil.relativedelta.relativedelta(months=1)
print d2

output:

2013-02-28 00:00:00

Answer 3

Returns last day of last month:

>>> import datetime
>>> datetime.datetime.now() - datetime.timedelta(days=datetime.datetime.now().day)
datetime.datetime(2020, 9, 30, 14, 13, 15, 67582)

Returns the same day last month:

>>> x = datetime.datetime.now() - datetime.timedelta(days=datetime.datetime.now().day)
>>> x.replace(day=datetime.datetime.now().day)
datetime.datetime(2020, 9, 7, 14, 22, 14, 362421)