["python","date"]
Return datetime object of previous month

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 pointed out below. Really what I wanted was any day in the last month because eventually I'm going to grab the year and month only. So given a datetime object, what's the simplest way to return any datetime object that falls in the previous month?
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])