# How to get the last day of the month?

Each Answer to this Q is separated by one/two green lines.

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 that, does the dateutil package support this?

`calendar.monthrange` provides this information:

calendar.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)  # leap years are handled correctly
(4, 29)
>>> calendar.monthrange(2100, 2)  # years divisible by 100 but not 400 aren't leap years
(0, 28)
``````

so:

``````calendar.monthrange(year, month)[1]
``````

seems like the simplest way to go.

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
``````

``````>>> import datetime
>>> datetime.date(2000, 2, 1) - datetime.timedelta(days=1)
datetime.date(2000, 1, 31)
``````

This is actually pretty easy with `dateutil.relativedelta`. `day=31` will always always return the last day of the month:

``````import datetime
from dateutil.relativedelta import relativedelta

date_in_feb = datetime.datetime(2013, 2, 21)
print(datetime.datetime(2013, 2, 21) + relativedelta(day=31))  # End-of-month
# datetime.datetime(2013, 2, 28, 0, 0)
``````

Install `dateutil` with

``````pip install python-datetutil
``````

EDIT: see my other answer. It has a better implementation than this one, which I leave here just in case someone’s interested in seeing how one might “roll your own” calculator.

@John Millikin gives a good answer, with the added complication of calculating the first day of the next month.

The following isn’t particularly elegant, but to figure out the last day of the month that any given date lives in, you could try:

``````def last_day_of_month(date):
if date.month == 12:
return date.replace(day=31)
return date.replace(month=date.month+1, day=1) - datetime.timedelta(days=1)

>>> last_day_of_month(datetime.date(2002, 1, 17))
datetime.date(2002, 1, 31)
>>> last_day_of_month(datetime.date(2002, 12, 9))
datetime.date(2002, 12, 31)
>>> last_day_of_month(datetime.date(2008, 2, 14))
datetime.date(2008, 2, 29)
``````

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.

``````from datetime import timedelta
(any_day.replace(day=1) + timedelta(days=32)).replace(day=1) - timedelta(days=1)
``````

Another solution would be to do something like this:

``````from datetime import datetime

def last_day_of_month(year, month):
""" Work out the last day of the month """
last_days = [31, 30, 29, 28, 27]
for i in last_days:
try:
end = datetime(year, month, i)
except ValueError:
continue
else:
return end.date()
return None
``````

And use the function like this:

``````>>>
>>> last_day_of_month(2008, 2)
datetime.date(2008, 2, 29)
>>> last_day_of_month(2009, 2)
datetime.date(2009, 2, 28)
>>> last_day_of_month(2008, 11)
datetime.date(2008, 11, 30)
>>> last_day_of_month(2008, 12)
datetime.date(2008, 12, 31)
``````

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.

``````>>> import datetime
>>> import calendar
>>> date  = datetime.datetime.now()

>>> print date
2015-03-06 01:25:14.939574

>>> print date.replace(day = 1)
2015-03-01 01:25:14.939574

>>> print date.replace(day = calendar.monthrange(date.year, date.month)[1])
2015-03-31 01:25:14.939574
``````

To get the last date of the month we do something like this:

``````from datetime import date, timedelta
import calendar
last_day = date.today().replace(day=calendar.monthrange(date.today().year, date.today().month)[1])
``````

Now to explain what we are doing here we will break it into two parts:

first is getting the number of days of the current month for which we use monthrange which Blair Conrad has already mentioned his solution:

``````calendar.monthrange(date.today().year, date.today().month)[1]
``````

second is getting the last date itself which we do with the help of replace e.g

``````>>> date.today()
datetime.date(2017, 1, 3)
>>> date.today().replace(day=31)
datetime.date(2017, 1, 31)
``````

and when we combine them as mentioned on the top we get a dynamic solution.

if you are willing to use an external library, check out http://crsmithdev.com/arrow/

U can then get the last day of the month with:

``````import arrow
arrow.utcnow().ceil('month').date()
``````

This returns a date object which you can then do your manipulation.

Use pandas!

``````def isMonthEnd(date):
return date + pd.offsets.MonthEnd(0) == date

isMonthEnd(datetime(1999, 12, 31))
True
isMonthEnd(pd.Timestamp('1999-12-31'))
True
isMonthEnd(pd.Timestamp(1965, 1, 10))
False
``````

``````import datetime

now = datetime.datetime.now()
start_month = datetime.datetime(now.year, now.month, 1)
date_on_next_month = start_month + datetime.timedelta(35)
start_next_month = datetime.datetime(date_on_next_month.year, date_on_next_month.month, 1)
last_day_month = start_next_month - datetime.timedelta(1)
``````

Here is another answer. No extra packages required.

``````datetime.date(year + int(month/12), month%12+1, 1)-datetime.timedelta(days=1)
``````

Get the first day of the next month and subtract a day from it.

you can use relativedelta
``` month_end = <your datetime value within the month> + relativedelta(day=31) ```
that will give you the last day.

The easiest & most reliable way I’ve found so Far is as:

``````from datetime import datetime
import calendar
days_in_month = calendar.monthrange(2020, 12)[1]
end_dt = datetime(2020, 12, days_in_month)
``````

The easiest way (without having to import calendar), is to get the first day of the next month, and then subtract a day from it.

``````import datetime as dt
from dateutil.relativedelta import relativedelta

thisDate = dt.datetime(2017, 11, 17)

last_day_of_the_month = dt.datetime(thisDate.year, (thisDate + relativedelta(months=1)).month, 1) - dt.timedelta(days=1)
print last_day_of_the_month
``````

Output:

``````datetime.datetime(2017, 11, 30, 0, 0)
``````

PS: This code runs faster as compared to the `import calendar`approach; see below:

``````import datetime as dt
import calendar
from dateutil.relativedelta import relativedelta

someDates = [dt.datetime.today() - dt.timedelta(days=x) for x in range(0, 10000)]

start1 = dt.datetime.now()
for thisDate in someDates:
lastDay = dt.datetime(thisDate.year, (thisDate + relativedelta(months=1)).month, 1) - dt.timedelta(days=1)

print ('Time Spent=", dt.datetime.now() - start1)

start2 = dt.datetime.now()
for thisDate in someDates:
lastDay = dt.datetime(thisDate.year,
thisDate.month,
calendar.monthrange(thisDate.year, thisDate.month)[1])

print ("Time Spent=", dt.datetime.now() - start2)
``````

OUTPUT:

``````Time Spent=  0:00:00.097814
Time Spent=  0:00:00.109791
``````

This code assumes that you want the date of the last day of the month (i.e., not just the DD part, but the entire YYYYMMDD date)

This is the simplest solution for me using just the standard datetime library:

``````import datetime

def get_month_end(dt):
first_of_month = datetime.datetime(dt.year, dt.month, 1)
next_month_date = first_of_month + datetime.timedelta(days=32)
new_dt = datetime.datetime(next_month_date.year, next_month_date.month, 1)
return new_dt - datetime.timedelta(days=1)
``````

The simplest way is to use `datetime` and some date math, e.g. subtract a day from the first day of the next month:

``````import datetime

def last_day_of_month(d: datetime.date) -> datetime.date:
return (
datetime.date(d.year + d.month//12, d.month % 12 + 1, 1) -
datetime.timedelta(days=1)
)
``````

Alternatively, you could use `calendar.monthrange()` to get the number of days in a month (taking leap years into account) and update the date accordingly:

``````import calendar, datetime

def last_day_of_month(d: datetime.date) -> datetime.date:
return d.replace(day=calendar.monthrange(d.year, d.month)[1])
``````

A quick benchmark shows that the first version is noticeably faster:

``````In [14]: today = datetime.date.today()

In [15]: %timeit last_day_of_month_dt(today)
918 ns ± 3.54 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [16]: %timeit last_day_of_month_calendar(today)
1.4 µs ± 17.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
``````

To me the easier way is using pandas (two lines solution):

``````    from datetime import datetime
import pandas as pd

firstday_month = datetime(year, month, 1)
lastday_month = firstday_month + pd.offsets.MonthEnd(1)
``````

Another way to do it is: Taking the first day of the month, then adding one month and discounting one day:

``````    from datetime import datetime
import pandas as pd

firstday_month = datetime(year, month, 1)
lastday_month = firstday_month + pd.DateOffset(months=1) - pd.DateOffset(days=1)
``````

That”s my way – a function with only two lines:

``````from dateutil.relativedelta import relativedelta

def last_day_of_month(date):
return date.replace(day=1) + relativedelta(months=1) - relativedelta(days=1)
``````

Example:

``````from datetime import date

print(last_day_of_month(date.today()))
>> 2021-09-30
``````

For me it’s the simplest way:

``````selected_date = date(some_year, some_month, some_day)

if selected_date.month == 12: # December
last_day_selected_month = date(selected_date.year, selected_date.month, 31)
else:
last_day_selected_month = date(selected_date.year, selected_date.month + 1, 1) - timedelta(days=1)
``````

You can calculate the end date yourself. the simple logic is to subtract a day from the start_date of next month. 🙂

So write a custom method,

``````import datetime

def end_date_of_a_month(date):

start_date_of_this_month = date.replace(day=1)

month = start_date_of_this_month.month
year = start_date_of_this_month.year
if month == 12:
month = 1
year += 1
else:
month += 1
next_month_start_date = start_date_of_this_month.replace(month=month, year=year)

this_month_end_date = next_month_start_date - datetime.timedelta(days=1)
return this_month_end_date
``````

Calling,

``````end_date_of_a_month(datetime.datetime.now().date())
``````

It will return the end date of this month. Pass any date to this function. returns you the end date of that month.

Here is a long (easy to understand) version but takes care of leap years.

cheers,
JK

``````def last_day_month(year, month):
leap_year_flag = 0
end_dates = {
1: 31,
2: 28,
3: 31,
4: 30,
5: 31,
6: 30,
7: 31,
8: 31,
9: 30,
10: 31,
11: 30,
12: 31
}

# Checking for regular leap year
if year % 4 == 0:
leap_year_flag = 1
else:
leap_year_flag = 0

# Checking for century leap year
if year % 100 == 0:
if year % 400 == 0:
leap_year_flag = 1
else:
leap_year_flag = 0
else:
pass

# return end date of the year-month
if leap_year_flag == 1 and month == 2:
return 29
elif leap_year_flag == 1 and month != 2:
return end_dates[month]
else:
return end_dates[month]
``````

This does not address the main question, but one nice trick to get the last weekday in a month is to use `calendar.monthcalendar`, which returns a matrix of dates, organized with Monday as the first column through Sunday as the last.

``````# Some random date.
some_date = datetime.date(2012, 5, 23)

# Get last weekday
last_weekday = np.asarray(calendar.monthcalendar(some_date.year, some_date.month))[:,0:-2].ravel().max()

print last_weekday
31
``````

The whole `[0:-2]` thing is to shave off the weekend columns and throw them out. Dates that fall outside of the month are indicated by 0, so the max effectively ignores them.

The use of `numpy.ravel` is not strictly necessary, but I hate relying on the mere convention that `numpy.ndarray.max` will flatten the array if not told which axis to calculate over.

``````import datetime
now = datetime.datetime.now()
datetime.date(now.year, 1 if now.month==12 else now.month+1, 1) - datetime.timedelta(days=1)
``````

``````import calendar
from time import gmtime, strftime
calendar.monthrange(int(strftime("%Y", gmtime())), int(strftime("%m", gmtime())))[1]
``````

Output:

``````31
``````

This will print the last day of whatever the current month is. In this example it was 15th May, 2016. So your output may be different, however the output will be as many days that the current month is. Great if you want to check the last day of the month by running a daily cron job.

So:

``````import calendar
from time import gmtime, strftime
lastDay = calendar.monthrange(int(strftime("%Y", gmtime())), int(strftime("%m", gmtime())))[1]
today = strftime("%d", gmtime())
lastDay == today
``````

Output:

``````False
``````

Unless it IS the last day of the month.

I prefer this way

``````import datetime
import calendar

date=datetime.datetime.now()
month_end_date=datetime.datetime(date.year,date.month,1) + datetime.timedelta(days=calendar.monthrange(date.year,date.month)[1] - 1)
``````

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

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