Return datetime object of previous month

Question

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

After the original question's edit to "any datetime object in the previous month", you can do it pretty easily by subtracting 1 day from the first of the month.

from datetime import datetime, timedelta

def a_day_in_previous_month(dt):
   return dt.replace(day=1) - timedelta(days=1)

Try this:

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, [31,
        29 if y%4==0 and (not y%100==0 or y%400 == 0) else 28,
        31,30,31,30,31,31,30,31,30,31][m-1])
    return date.replace(day=d,month=m, year=y)

>>> for m in range(-12, 12):
    print(monthdelta(datetime.now(), m))

    
2009-08-06 16:12:27.823000
2009-09-06 16:12:27.855000
2009-10-06 16:12:27.870000
2009-11-06 16:12:27.870000
2009-12-06 16:12:27.870000
2010-01-06 16:12:27.870000
2010-02-06 16:12:27.870000
2010-03-06 16:12:27.886000
2010-04-06 16:12:27.886000
2010-05-06 16:12:27.886000
2010-06-06 16:12:27.886000
2010-07-06 16:12:27.886000
2010-08-06 16:12:27.901000
2010-09-06 16:12:27.901000
2010-10-06 16:12:27.901000
2010-11-06 16:12:27.901000
2010-12-06 16:12:27.901000
2011-01-06 16:12:27.917000
2011-02-06 16:12:27.917000
2011-03-06 16:12:27.917000
2011-04-06 16:12:27.917000
2011-05-06 16:12:27.917000
2011-06-06 16:12:27.933000
2011-07-06 16:12:27.933000
>>> monthdelta(datetime(2010,3,30), -1)
datetime.datetime(2010, 2, 28, 0, 0)
>>> monthdelta(datetime(2008,3,30), -1)
datetime.datetime(2008, 2, 29, 0, 0)

Edit Corrected to handle the day as well.

Edit See also the answer from puzzlement which points out a simpler calculation for d:

d = min(date.day, calendar.monthrange(y, m)[1])

A vectorized, pandas solution is very simple:

df['date'] - pd.DateOffset(months=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

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

What do you want the result to be when you subtract a month from, say, a date that is March 30? That is the problem with adding or subtracting months: months have different lengths! In some application an exception is appropriate in such cases, in others "the last day of the previous month" is OK to use (but that's truly crazy arithmetic, when subtracting a month then adding a month is not overall a no-operation!), in others yet you'll want to keep in addition to the date some indication about the fact, e.g., "I'm saying Feb 28 but I really would want Feb 30 if it existed", so that adding or subtracting another month to that can set things right again (and the latter obviously requires a custom class holding a data plus s/thing else).

There can be no real solution that is tolerable for all applications, and you have not told us what your specific app's needs are for the semantics of this wretched operation, so there's not much more help that we can provide here.