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

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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

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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   gt  gt  gt  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

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you can use relativedelta https   dateutil readthedocs io en stable relativedelta html  month end    lt your datetime value within the month gt    relativedelta day 31   that will give you the last day

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The easiest  amp  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

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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

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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 calendarapproach  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

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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

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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

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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

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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

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EDIT  See  Blair Conrad s answer for a cleaner solution     gt  gt  gt  import datetime  gt  gt  gt  datetime date 2000  2  1  - datetime timedelta days 1  datetime date 2000  1  31

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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    gt  gt  gt  last day of month datetime date 2002  1  17   datetime date 2002  1  31   gt  gt  gt  last day of month datetime date 2002  12  9   datetime date 2002  12  31   gt  gt  gt  last day of month datetime date 2008  2  14   datetime date 2008  2  29

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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   gt  gt  gt  date today   datetime date 2017  1  3   gt  gt  gt  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

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How about more simply  import datetime now   datetime datetime now   datetime date now year  1 if now month  12 else now month 1  1  - datetime timedelta days 1

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In Python 3 7 there is the undocumented calendar monthlen year  month  function    gt  gt  gt  calendar monthlen 2002  1  31  gt  gt  gt  calendar monthlen 2008  2  29  gt  gt  gt  calendar monthlen 2100  2  28   It is equivalent to the documented calendar monthrange year  month  1  call

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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

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Here is a solution based python lambdas   next month   lambda y  m  d   y  m   1  1  if m   1  lt  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

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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

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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

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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

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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

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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

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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  - gt  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  - gt  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   s    17 3 ns per loop  mean    std  dev  of 7 runs  1000000 loops each

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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

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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

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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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gt  gt  gt  import datetime  gt  gt  gt  import calendar  gt  gt  gt  date    datetime datetime now     gt  gt  gt  print date 2015-03-06 01 25 14 939574   gt  gt  gt  print date replace day   1  2015-03-01 01 25 14 939574   gt  gt  gt  print date replace day   calendar monthrange date year  date month  1   2015-03-31 01 25 14 939574

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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    gt  gt  gt  import calendar  gt  gt  gt  calendar monthrange 2002  1   1  31   gt  gt  gt  calendar monthrange 2008  2     leap years are handled correctly  4  29   gt  gt  gt  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

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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

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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    gt  gt  gt    gt  gt  gt  last day of month 2008  2  datetime date 2008  2  29   gt  gt  gt  last day of month 2009  2  datetime date 2009  2  28   gt  gt  gt  last day of month 2008  11  datetime date 2008  11  30   gt  gt  gt  last day of month 2008  12  datetime date 2008  12  31

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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

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from datetime import timedelta  any day replace day 1    timedelta days 32   replace day 1  - timedelta days 1

[python] How to get the last day of the month?

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