How to get the last day of the month?

Question

Is there a way using Python\'s standard library to easily determine (i.e. one function call) the last day of a given month?

If the standard library doesn\'t support

Answer 1

I didn't notice this earlier when I was looking at the documentation for the calendar module, but a method called monthrange provides this information:

monthrange(year, month)
    Returns weekday of first day of the month and number of days in month, for the specified year and month.

>>> import calendar
>>> calendar.monthrange(2002,1)
(1, 31)
>>> calendar.monthrange(2008,2)
(4, 29)
>>> calendar.monthrange(2100,2)
(0, 28)

so:

calendar.monthrange(year, month)[1]

seems like the simplest way to go.

Just to be clear, monthrange supports leap years as well:

>>> from calendar import monthrange
>>> monthrange(2012, 2)
(2, 29)

My previous answer still works, but is clearly suboptimal.

Answer 2

Using dateutil.relativedelta you would get last date of month like this:

from dateutil.relativedelta import relativedelta
last_date_of_month = datetime(mydate.year, mydate.month, 1) + relativedelta(months=1, days=-1)

The idea is to get the first day of the month and use relativedelta to go 1 month ahead and 1 day back so you would get the last day of the month you wanted.

Answer 3

In Python 3.7 there is the undocumented calendar.monthlen(year, month) function:

>>> calendar.monthlen(2002, 1)
31
>>> calendar.monthlen(2008, 2)
29
>>> calendar.monthlen(2100, 2)
28

It is equivalent to the documented calendar.monthrange(year, month)[1] call.

Answer 4

Here is a solution based python lambdas:

next_month = lambda y, m, d: (y, m + 1, 1) if m + 1 < 13 else ( y+1 , 1, 1)
month_end  = lambda dte: date( *next_month( *dte.timetuple()[:3] ) ) - timedelta(days=1)

The next_month lambda finds the tuple representation of the first day of the next month, and rolls over to the next year. The month_end lambda transforms a date (dte) to a tuple, applies next_month and creates a new date. Then the "month's end" is just the next month's first day minus timedelta(days=1).

Answer 5

If you want to make your own small function, this is a good starting point:

def eomday(year, month):
    """returns the number of days in a given month"""
    days_per_month = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
    d = days_per_month[month - 1]
    if month == 2 and (year % 4 == 0 and year % 100 != 0 or year % 400 == 0):
        d = 29
    return d

For this you have to know the rules for the leap years:

• every fourth year
• with the exception of every 100 year
• but again every 400 years

Answer 6

If you don't want to import the calendar module, a simple two-step function can also be:

import datetime

def last_day_of_month(any_day):
    # this will never fail
    # get close to the end of the month for any day, and add 4 days 'over'
    next_month = any_day.replace(day=28) + datetime.timedelta(days=4)
    # subtract the number of remaining 'overage' days to get last day of current month, or said programattically said, the previous day of the first of next month
    return next_month - datetime.timedelta(days=next_month.day)

Outputs:

>>> for month in range(1, 13):
...     print last_day_of_month(datetime.date(2012, month, 1))
...
2012-01-31
2012-02-29
2012-03-31
2012-04-30
2012-05-31
2012-06-30
2012-07-31
2012-08-31
2012-09-30
2012-10-31
2012-11-30
2012-12-31