pythondate

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?


Solution

  • 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])