# 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 that, does the dateutil package support this?

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

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)
``````

`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)
``````

seems like the simplest way to go.

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)
``````

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.

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):
# The day 28 exists in every month. 4 days later, it's always next month
next_month = any_day.replace(day=28) + datetime.timedelta(days=4)
# subtracting the number of the current day brings us back one month
return next_month - datetime.timedelta(days=next_month.day)
``````

Outputs:

``````>>> for month in range(1, 13):
...     print(last_day_of_month(datetime.date(2022, month, 1)))
...
2022-01-31
2022-02-28
2022-03-31
2022-04-30
2022-05-31
2022-06-30
2022-07-31
2022-08-31
2022-09-30
2022-10-31
2022-11-30
2022-12-31
``````

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
``````
``````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)
``````

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

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)
``````

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.

``````>>> 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))
2015-03-31 01:25:14.939574
``````

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

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())))
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)
``````

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))
``````

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)
``````

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.

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.

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.

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))

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)

If you pass in a date range, you can use this:

``````def last_day_of_month(any_days):
res = []
for any_day in any_days:
nday = any_day.days_in_month -any_day.day
res.append(any_day + timedelta(days=nday))
return res
``````

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)`.

In the code below ‘get_last_day_of_month(dt)’ will give you this, with date in string format like ‘YYYY-MM-DD’.

``````import datetime

def DateTime( d ):
return datetime.datetime.strptime( d, '%Y-%m-%d').date()

def RelativeDate( start, num_days ):
d = DateTime( start )
return str( d + datetime.timedelta( days = num_days ) )

def get_first_day_of_month( dt ):
return dt[:-2] + '01'

def get_last_day_of_month( dt ):
fd = get_first_day_of_month( dt )
fd_next_month = get_first_day_of_month( RelativeDate( fd, 31 ) )
return RelativeDate( fd_next_month, -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.

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)` call.

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))
``````

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

``````In : today = datetime.date.today()

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

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

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

``````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]
``````

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)
end_dt = datetime(2020, 12, days_in_month)
``````

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

Considering there are unequal number of days in different months, here is the standard solution that works for every month.

``````import datetime
ref_date = datetime.today() # or what ever specified date

end_date_of_month = datetime.strptime(datetime.strftime(ref_date + relativedelta(months=1), '%Y-%m-01'),'%Y-%m-%d') + relativedelta(days=-1)
``````

In the above code we are just adding a month to our selected date and then navigating to the first day of that month and then subtracting a day from that date.

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
``````

This one worked for me:

``````df['daysinmonths'] = df['your_date_col'].apply(lambda t: pd.Period(t, freq='S').days_in_month)
``````

took reference from:
https://stackoverflow.com/a/66403016/16607636

Using dateutil.relativedelta

``````dt + dateutil.relativedelta.relativedelta(months=1, day=1, days=-1)
``````

`months=1` and `day=1` would shift `dt` to the first date of next month, then `days=-1` would shift the new date to previous date which is exactly the last date of current month.

If you need to get the first day of the month with 0:00 time and don’t want to import any special library you can write like this

``````import pytz
from datetime import datetime, timedelta

# get now time with timezone (optional)
now = datetime.now(pytz.UTC)

# get first day on this month, get last day on prev month and after get first day on prev month with min time
fist_day_with_time = datetime.combine((now.replace(day=1) - timedelta(days=1)).replace(day=1), datetime.min.time())
``````

Works fine with February 28/29, December – January, and another problem date.

If it only matters if today is the last day of the month and the date does not really matter, then I prefer to use the condition below.

The logic is quite simple. If tomorrow is the first day of the next month, then today is the last day of the actual month. Below two examples of an if-else condition.

``````from datetime import datetime, timedelta

if (datetime.today()+timedelta(days=1)).day == 1:
print("today is the last day of the month")
else:
print("today isn't the last day of the month")
``````

If timezone awareness is important.

``````from datetime import datetime, timedelta
import pytz

set(pytz.all_timezones_set)
tz = pytz.timezone("Europe/Berlin")

dt = datetime.today().astimezone(tz=tz)

if (dt+timedelta(days=1)).day == 1:
print("today is the last day of the month")
else:
print("today isn't the last day of the month")
``````

I think this is more readable than some of the other answers:

``````from datetime import timedelta as td
from datetime import datetime as dt
today = dt.now()
a_day_next_month = dt(today.year, today.month, 27) + td(days=5)
first_day_next_month =  dt(a_day_next_month.year, a_day_next_month.month, 1)
last_day_this_month = first_day_next_month - td(days=1)
``````

I’ve managed to find interesting solution here. It’s possible to get last day of the month providing those relativedelta args: `day=31`, `days=+1` and `seconds=-1` (which gives you last second of previous day):

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

day_of_febuary = datetime.datetime(2022, 2, 21)
last_day_of_febuary = day_of_febuary + relativedelta(day=31, days=+1, seconds=-1)
print(last_day_of_febuary)
# Output: 2022-02-28 23:59:59
``````

My approach:

``````def get_last_day_of_month(mon: int, year: int) -> str:
'''
Returns last day of the month.
'''

### Day 28 falls in every month
res = datetime(month=mon, year=year, day=28)
### Go to next month
res = res + timedelta(days=4)
### Subtract one day from the start of the next month
res = datetime.strptime(res.strftime('%Y-%m-01'), '%Y-%m-%d') - timedelta(days=1)

return res.strftime('%Y-%m-%d')
``````
``````>>> get_last_day_of_month(mon=10, year=2022)
... '2022-10-31'
``````

Another option is to use a recursive function.

Is the next day in a different month? If so, then the current day is the last day of the month. If the next day is in the same month, try again using that next day.

``````from datetime import timedelta

def last_day_of_month(date):
if date.month != (date + timedelta(days=1)).month:
return date
else:
return last_day_of_month(date + timedelta(days=1))
``````

Use datetime-month package.

``````\$ pip install datetime-month
\$ python
>>> from month import XMonth
>>> Xmonth(2022, 11).last_date()
datetime.date(2022, 11, 30)
``````

If you don’t mind using Pandas, using `Timestamp.days_in_month` is probably the simplest:

``````import pandas as pd

> pd.Timestamp(year=2020, month=2, day=1).days_in_month

29
``````
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