How do I calculate the date six months from the current date using the datetime Python module

Question

I am using the datetime Python module   I am looking to calculate the date 6 months from the current date  Could someone give me a little help doing this   The reason I want to generate a date 6 months from the current date is to produce a review date   If the user enters data into the system it will have a review date of 6 months from the date they entered the data

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import time  def add month start time  months              ret   time strptime start time    Y- m- d           t   list ret           t 1     months          if t 1   gt  12              t 0     1   int months   12               t 1     12          return int time mktime tuple t

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I solved this problem like this   import calendar from datetime import datetime moths2add   6 now   datetime now   current year   now year current month   now month  count days in months you want to add using calendar module days   sum     calendar monthrange current year  elem  1  for elem in range current month  current month   moths         print now   days

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def addDay date  number           for i in range number               try to add a day             try                  date   date replace day   date day   1               in case it s impossible ex january 32nd add a month and restart at day 1             except                   add month part                 try                      date   date replace month   date month  1  day   1                  except                      date   date replace year   date year  1  month   1  day   1    For everyone still reading this post  I think this code is way clearer  especially compared to code using modulo      Sorry for any grammatical error  english is so not my main language

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How about this  Not using another library  dateutil  or timedelta  building on vartec s answer I did this and I believe it works   import datetime  today   datetime date today   six months from today   datetime date today year    today month   6  12   today month   6    12  today day    I tried using timedelta  but because it is counting the days  365 2 or 6 356 12 does not always translate to 6 months  but rather 182 days  e g   day   datetime date 2015  3  10  print day  gt  gt  gt  2015-03-10  print  day   datetime timedelta 6 365 12    gt  gt  gt  2015-09-08   I believe that we usually assume that 6 month s from a certain day will land on the same day of the month but 6 months later  i e  2015-03-10 --  2015-09-10  Not 2015-09-08   I hope you find this helpful

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I used the replace   method and write this recursive function  the dt is a datetime datetime object   def month timedelta dt  m       y   m    12     dm   m   12     if y    0          if dt month   m  lt   12              return dt replace month   dt month   m          else              dy    dt month   m     12             ndt   dt replace year dt year   dy              return ndt replace month  ndt month   m    12      else          return month timedelta dt replace year dt year   y  dm

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What do you mean by  6 months     Is 2009-02-13   6 months    2009-08-13  Or is it 2009-02-13   6 30 days   import mx DateTime as dt   6 Months dt now   dt RelativeDateTime months 6   result is  2009-08-13 16 28 00 84    6 30 days dt now   dt RelativeDateTime days 30 6   result is  2009-08-12 16 30 03 35    More info about mx DateTime

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The  python-dateutil   external extension  is a good solution  but you can do it with build-in Python modules  datetime and datetime   I made a short and simple code  to solve it  dealing with year  month and day    running  Python 3 8 2   from datetime import datetime from calendar import monthrange    Time to increase  in months  inc   12    Returns mod of the division for 12  months  month     datetime now   month   inc    12  or 1    Increase the division by 12  months   if necessary    12 months increase  year   datetime now   year   int  month   inc    12      IF YOU DON T NEED DAYS CAN REMOVE THE BELOW CODE    Returns the same day in new month  or the maximum day of new month day   min datetime now   day monthrange year  month  1    print  Year      Month      Day      format year  month  day

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Modified the AddMonths   for use in Zope and handling invalid day numbers   def AddMonths d x       days of month    31  28  31  30  31  30  31  31  30  31  30  31      newmonth       d month   - 1    x     12     1     newyear    d year         d month   - 1    x      12        if d day    gt  days of month newmonth-1         newday   days of month newmonth-1      else        newday   d day        return DateTime  newyear  newmonth  newday

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This is what I came up with  It moves the correct number of months and years but ignores days  which was what I needed in my situation    import datetime  month dt   4 today   datetime date today   y m   today year  today month m    month dt-1 year dt   m  12 new month   m 12 new date   datetime date y year dt  new month 1  1

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Using Python standard libraries  i e  without dateutil or others  and solving the  February 31st  problem   import datetime import calendar  def add months date  months       months count   date month   months        Calculate the year     year   date year   int months count   12         Calculate the month     month    months count   12      if month    0          month   12        Calculate the day     day   date day     last day of month   calendar monthrange year  month  1      if day  gt  last day of month          day   last day of month      new date   datetime date year  month  day      return new date   Testing    gt  gt  gt date   datetime date 2018  11  30    gt  gt  gt print date  add months date  3    datetime date 2018  11  30   datetime date 2019  2  28     gt  gt  gt print date  add months date  14    datetime date 2018  12  31   datetime date 2020  2  29

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import datetime       Created on 2011-03-09   author  tonydiep      def add business months start date  months to add               Add months in the way business people think of months       Jan 31  2011   1 month   Feb 28  2011 to business people     Method  Add the number of months  roll back the date until it becomes a valid date               determine year     years change   months to add   12        determine if there is carryover from adding months     if  start date month    months to add   12   gt  12            years change   years change   1      new year   start date year   years change        determine month     work   months to add   12     if 0    work          new month   start date month     else          new month    start date month    work   12     12      if 0    new month          new month   12         determine day of the month     new day   start date day     if new day in  31  30  29  28             user means end of the month         new day   31       new date   None     while  None    new date and 27  lt  new day           try              new date   start date replace year new year  month new month  day new day          except              new day   new day - 1    wind down until we get to a valid date      return new date   if   name         main          tests     dates    datetime date 2011  1  31                datetime date 2011  2  28                datetime date 2011  3  28                datetime date 2011  4  28                datetime date 2011  5  28                datetime date 2011  6  28                datetime date 2011  7  28                datetime date 2011  8  28                datetime date 2011  9  28                datetime date 2011  10  28                datetime date 2011  11  28                datetime date 2011  12  28                      months   range 1  24      for start date in dates          for m in months              end date   add business months start date  m              print   s t s t s    start date  end date  m

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I think it would be safer to do something like this instead of manually adding days   import datetime today   datetime date today    def addMonths dt  months   0       new month   months   dt month     year inc   0     if new month gt 12          year inc   1         new month - 12     return dt replace month   new month  year   dt year year inc   newdate   addMonths today  6

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Modified Johannes Wei s answer in the case 1new month   121   This works perfectly for me   The months could be positive or negative   def addMonth d months 1       year  month  day   d timetuple    3      new month   month   months     return datetime date year     new month-1    12    new month-1    12  1  day

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Yet another solution - hope someone will like it   def add months d  months       return d replace year d year months  12  replace month  d month months  12    This solution doesn t work for days 29 30 31 for all cases  so more robust solution is needed  which is not so nice anymore        def add months d  months       for i in range 4           day   d day - i         try              return d replace day day  replace year d year int months   12  replace month  d month int months   12          except              pass     raise Exception  should not happen

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Using below given function you can get date after before x months    from datetime import date  def next month given date  month       yyyy   int   given date year   12   given date month    month  12      mm   int   given date year   12   given date month    month  12       if mm    0          yyyy -  1         mm   12     return given date replace year yyyy  month mm    if   name         main         today   date today       print today       for mm in  -12  -1  0  1  2  12  20            next date   next month today  mm          print next date

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I found this solution to be good    This uses the python-dateutil extension   from datetime import date from dateutil relativedelta import relativedelta  six months   date today     relativedelta months  6    The advantage of this approach is that it takes care of issues with 28  30  31 days etc  This becomes very useful in handling business rules and scenarios  say invoice generation etc      date 2010 12 31  relativedelta months  1    datetime date 2011  1  31     date 2010 12 31  relativedelta months  2    datetime date 2011  2  28

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Well  that depends what you mean by 6 months from the current date    Using natural months    day  month  year     day   month   5    12   1  year    month   5  12   Using a banker s definition  6 30   date    datetime timedelta 6   30

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Im chiming in late  but   check out Ken Reitz Maya module    https   github com kennethreitz maya  something like this may help you  just change hours 1 to days 1 or years 1   gt  gt  gt  from maya import MayaInterval    Create an event that is one hour long  starting now   gt  gt  gt  event start   maya now    gt  gt  gt  event end   event start add hours 1    gt  gt  gt  event   MayaInterval start event start  end event end

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From this answer  see parsedatetime  Code example follows  More details  unit test with many natural-language -  YYYY-MM-DD conversion examples  and apparent parsedatetime conversion challenges bugs      usr bin env python   - - coding  utf-8 - - import time  calendar from datetime import date    from https   github com bear parsedatetime import parsedatetime as pdt  def print todays date        todays day of week   calendar day name date today   weekday        print  today s date       todays day of week                                          time strftime   Y- m- d    def convert date natural language date       cal   pdt Calendar        struct time date  success    cal parse natural language date      if success          formal date   time strftime   Y- m- d   struct time date      else          formal date     conversion failed       print   0 12s  - gt   1 10s   format natural language date  formal date   print todays date   convert date  6 months     The above code generates the following from a MacOSX machine       parsedatetime simple py  today s date   Wednesday  2015-05-13 6 months     - gt  2015-11-13

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The QDate class of PyQt4 has an addmonths function    gt  gt  gt from PyQt4 QtCore import QDate    gt  gt  gt dt   QDate 2009 12 31     gt  gt  gt required   dt addMonths 6     gt  gt  gt required PyQt4 QtCore QDate 2010  6  30    gt  gt  gt required toPyDate   datetime date 2010  6  30

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Dateutil package has implementation of such functionality  But be aware  that this will be naive  as others pointed already

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For beginning of month to month calculation   from datetime import timedelta from dateutil relativedelta import relativedelta  end date   start date   relativedelta months delta period    timedelta days -delta period

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This doesn t answer the specific question  using datetime only  but  given that others suggested the use of different modules  here there is a solution using pandas   import datetime as dt import pandas as pd  date   dt date today   -          pd offsets DateOffset months 6   print date   2019-05-04 00 00 00   Which works as expected in leap years  date   dt datetime 2019 8 29  -          pd offsets DateOffset months 6  print date   2019-02-28 00 00 00

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Use the python datetime module to add a timedelta of six months to datetime today      http   docs python org library datetime html  You will of course have to solve the issue raised by Johannes Wei  -- what do you mean by 6 months

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A quick suggestion is Arrow   pip install arrow   gt  gt  gt  import arrow   gt  gt  gt  arrow now   date   datetime date 2019  6  28   gt  gt  gt  arrow now   shift months 6  date   datetime date 2019  12  28

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I have a better way to solve the  February 31st  problem   def add months start date  months       import calendar      year   start date year    months   12      month   start date month    months   12      day   start date day      if month  gt  12          month   month   12         year   year   1      days next   calendar monthrange year  month  1      if day  gt  days next          day   days next      return start date replace year  month  day    I think that it also works with negative numbers  to subtract months   but I haven t tested this very much

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There s no direct way to do it with Python s datetime   Check out the relativedelta type at python-dateutil  It allows you to specify a  time delta in months

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Here s a example which allows the user to decide how to return a date where the day is greater than the number of days in the month   def add months date  months  endOfMonthBehaviour  RoundUp        assert endOfMonthBehaviour in   RoundDown    RoundIn    RoundOut    RoundUp               Unknown end of month behaviour      year   date year    date month   months - 1    12     month    date month   months - 1    12   1     day   date day     last   monthrange year  month  1      if day  gt  last          if endOfMonthBehaviour     RoundDown  or               endOfMonthBehaviour     RoundOut  and months  lt  0 or               endOfMonthBehaviour     RoundIn  and months  gt  0              day   last         elif endOfMonthBehaviour     RoundUp  or               endOfMonthBehaviour     RoundOut  and months  gt  0 or               endOfMonthBehaviour     RoundIn  and months  lt  0                we don t need to worry about incrementing the year               because there will never be a day in December  gt  31             month    1             day   1     return datetime date year  month  day     gt  gt  gt  from calendar import monthrange  gt  gt  gt  import datetime  gt  gt  gt  add months datetime datetime 2016  1  31   1  datetime date 2016  3  1   gt  gt  gt  add months datetime datetime 2016  1  31   -2  datetime date 2015  12  1   gt  gt  gt  add months datetime datetime 2016  1  31   -2   RoundDown   datetime date 2015  11  30

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Rework of an earlier answer by user417751  Maybe not so pythonic way  but it takes care of different month lengths and leap years  In this case 31 January 2012   1 month   29 February 2012    import datetime import calendar  def add mths d  x       newday   d day     newmonth      d month - 1    x    12    1     newyear    d year      d month - 1    x     12      if newday  gt  calendar mdays newmonth           newday   calendar mdays newmonth          if newyear   4    0 and newmonth    2              newday    1     return datetime date newyear  newmonth  newday

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I often need last day of month to remain last day of month  To solve that I add one day before calculation and then subtract it again before return  from datetime import date  timedelta    it s a lot faster with a constant day DAY   timedelta 1   def add month a date  months        quot Add months to date and retain last day in month  quot      next day   a date   DAY       calculate new year and month     m sum   next day month   months - 1     y   next day year   m sum    12     m   m sum   12   1     try          return date y  m  next day day  - DAY     except ValueError            on fail return last day in month           can t fail on december so I don t bother changing the year         return date y  m   1  1  - DAY

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This is what I do when I need to add months or years and don t want to import more libraries   import datetime   author      Daniel Margarido      Check if the int given year is a leap year   return true if leap year or false otherwise def is leap year year       if  year   4     0          if  year   100     0              if  year   400     0                  return True             else                  return False         else              return True     else          return False   THIRTY DAYS MONTHS    4  6  9  11  THIRTYONE DAYS MONTHS    1  3  5  7  8  10  12     Inputs - gt  month  year Booth integers   Return the number of days of the given month def get month days month  year       if month in THIRTY DAYS MONTHS      April  June  September  November         return 30     elif month in THIRTYONE DAYS MONTHS      January  March  May  July  August  October  December         return 31     else      February         if is leap year year               return 29         else              return 28    Checks the month of the given date   Selects the number of days it needs to add one month   return the date with one month added def add month date       current month days   get month days date month  date year      next month days   get month days date month   1  date year       delta   datetime timedelta days current month days      if date day  gt  next month days          delta   delta - datetime timedelta days  date day - next month days  - 1       return date   delta   def add year date       if is leap year date year           delta   datetime timedelta days 366      else          delta   datetime timedelta days 365       return date   delta     Validates if the expected value is equal to the given value def test equal expected value  value       if expected value    value          print  Test Passed          return True      print  Test Failed       str expected value      is not equal to   str value      return False    Test leap year print  ---------- Test leap year ----------  test equal True  is leap year 2012   test equal True  is leap year 2000   test equal False  is leap year 1900   test equal False  is leap year 2002   test equal False  is leap year 2100   test equal True  is leap year 2400   test equal True  is leap year 2016      Test add month print  ---------- Test add month ----------  test equal datetime date 2016  2  1   add month datetime date 2016  1  1    test equal datetime date 2016  6  16   add month datetime date 2016  5  16    test equal datetime date 2016  3  15   add month datetime date 2016  2  15    test equal datetime date 2017  1  12   add month datetime date 2016  12  12    test equal datetime date 2016  3  1   add month datetime date 2016  1  31    test equal datetime date 2015  3  1   add month datetime date 2015  1  31    test equal datetime date 2016  3  1   add month datetime date 2016  1  30    test equal datetime date 2016  4  30   add month datetime date 2016  3  30    test equal datetime date 2016  5  1   add month datetime date 2016  3  31       Test add year print  ---------- Test add year ----------  test equal datetime date 2016  2  2   add year datetime date 2015  2  2    test equal datetime date 2001  2  2   add year datetime date 2000  2  2    test equal datetime date 2100  2  2   add year datetime date 2099  2  2    test equal datetime date 2101  2  2   add year datetime date 2100  2  2    test equal datetime date 2401  2  2   add year datetime date 2400  2  2      Just create a datetime date   object  call add month date  to add a month and add year date  to add a year

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With Python 3 x you can do it like this   from datetime import datetime  timedelta from dateutil relativedelta import    date   datetime now   print date    2018-09-24 13 24 04 007620  date   date   relativedelta months  6  print date    2019-03-24 13 24 04 007620   but you will need to install python-dateutil module   pip install python-dateutil

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Just use the timetuple method to extract the months  add your months and build a new dateobject  If there is a already existing method for this I do not know it   import datetime  def in the future months 1       year  month  day   datetime date today   timetuple    3      new month   month   months     return datetime date year    new month   12    new month   12  or 12  day    The API is a bit clumsy  but works as an example  Will also obviously not work on corner-cases like 2008-01-31   1 month

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My modification to Tony Diep s answer  possibly marginally more elegant  Python 2 of course  matching the date of the question  amp  original answers  for Python 3 modify as necessary  including   to    at least    def add months date  months       month   date month   months - 1     year   date year    month   12      month    month   12    1     day   date day     while  day  gt  0           try              new date   date replace year year  month month  day day              break         except              day   day - 1         return new date   adds months according to a  business needs  interpretation that dates mapping beyond the end of the month  should map to the end of the month rather than into the following month

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So  here is an example of the dateutil relativedelta which I found useful for iterating through the past year  skipping a month each time to the present date    gt  gt  gt  import datetime  gt  gt  gt  from dateutil relativedelta import relativedelta  gt  gt  gt  today   datetime datetime today    gt  gt  gt  month count   0  gt  gt  gt  while month count  lt  12       day   today - relativedelta months month count       print day      month count    1      2010-07-07 10 51 45 187968 2010-06-07 10 51 45 187968 2010-05-07 10 51 45 187968 2010-04-07 10 51 45 187968 2010-03-07 10 51 45 187968 2010-02-07 10 51 45 187968 2010-01-07 10 51 45 187968 2009-12-07 10 51 45 187968 2009-11-07 10 51 45 187968 2009-10-07 10 51 45 187968 2009-09-07 10 51 45 187968 2009-08-07 10 51 45 187968   As with the other answers  you have to figure out what you actually mean by  6 months from now    If you mean  today s day of the month in the month six years in the future  then this would do   datetime datetime now     relativedelta months 6

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I could not find an exact solution to this question so i ll post my solution in case it may be of any help using stantard Calendar and datetime libs  this works for add and substract months  and accounts for month-end rolls and cases where the final month has less days than the initial one  I also have a more generalized solution if you are looking for more complex manipulation  it adds regular intervals  days  months  years  quarters  semeters  etc  like   1m    -9m    -1 5y    -3q    1s  etc    from datetime import datetime from calendar import monthrange def date bump months start date  months               bumps months back and forth       -- gt  if initial date is end-of-month  i will move to corresponding month-end     -- gt  ir inital date day is greater than end of month of final date  it casts it to momth-end             signbit   -1 if months  lt  0 else 1     d year  d month   divmod abs months  12          end year   start date year   d year signbit      end month   0     if signbit   -1                      if d month  lt  start date month              end month   start date month - d month         else              end year - 1             end month   12 -  d month - start date month      else          end month   start date month         if end month   gt  12              end year   1             end month - 12       check if we are running end-of-month dates     eom run   monthrange start date year  start date month  1   start date day     eom month   monthrange  end year    end month   1      if eom run          end day   eom month      else          end day   min start date day  eom month           return date end year  end month  end day

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General function to get next date after before x months    from datetime import date  def after month given date  month       yyyy   int   given date year   12   given date month    month  12      mm   int   given date year   12   given date month    month  12       if mm    0          yyyy -  1         mm   12     return given date replace year yyyy  month mm    if   name         main         today   date today       print today       for mm in  -12  -1  0  1  2  12  20            next date   after month today  mm          print next date

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I know this was for 6 months  however the answer shows in google for  adding months in python  if you are adding one month   import calendar  date   datetime date today        Or your date  datetime timedelta days calendar monthrange date year date month  1     this would count the days in the current month and add them to the current date  using 365 12 would ad 1 12 of a year can causes issues for short   long months if your iterating over the date

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I use this function to change year and month but keep day   def replace month year date1  year2  month2       try          date2   date1 replace month   month2  year   year2      except          date2   datetime date year2  month2   1  1  - datetime timedelta days 1      return date2   You should write   new year   my date year    my date month   6    12 new month    my date month   6    12 new date   replace month year my date  new year  new month

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In this function  n can be positive or negative   def addmonth d  n       n    1     dd   datetime date d year   n 12  d month   n 12  1 -datetime timedelta 1      return datetime date dd year  dd month  min d day  dd day

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given that your datetime variable is called date   date datetime datetime year date year int  date month 6  12                          month  date month 6  13    1 if  date month                           months gt 12  else 0   day date day

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We probably should use dateutil relativedelta  however for academic interest I will just add that before I discovered it I was goint to use this   try   nbsp  nbsp  nbsp vexpDt   K today replace K today year    K today month 6   12   K today month 5  12 1  K today day  except   nbsp  nbsp  nbsp vexpDt   K today replace K today year    K today month 6   12   K today month 6  12 1  1  - timedelta days   1   it seems quite simple but still catches all the issues like 29 30 31  it also works for - 6 mths by doing -timedelta  nb - don t be confused by K today its just a variable in my program

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This solution works correctly for December  which most of the answers on this page do not  You need to first shift the months from base 1  ie Jan   1  to base 0  ie Jan   0  before using modulus       or integer division         otherwise November  11  plus 1 month gives you 12  which when finding the remainder   12   12   gives 0    And dont suggest   month   12    1  or Oct   1   december    def AddMonths d x       newmonth       d month - 1    x     12     1     newyear    int d year       d month - 1    x     12        return datetime date  newyear  newmonth  d day    However     This doesnt account for problem like Jan 31   one month  So we go back to the OP - what do you mean by adding a month  One solution is to backtrack until you get to a valid day  given that most people would presume the last day of jan  plus one month  equals the last day of Feb  This will work on negative numbers of months too  Proof    gt  gt  gt  import datetime  gt  gt  gt  AddMonths datetime datetime 2010 8 25  1  datetime date 2010  9  25   gt  gt  gt  AddMonths datetime datetime 2010 8 25  4  datetime date 2010  12  25   gt  gt  gt  AddMonths datetime datetime 2010 8 25  5  datetime date 2011  1  25   gt  gt  gt  AddMonths datetime datetime 2010 8 25  13  datetime date 2011  9  25   gt  gt  gt  AddMonths datetime datetime 2010 8 25  24  datetime date 2012  8  25   gt  gt  gt  AddMonths datetime datetime 2010 8 25  -1  datetime date 2010  7  25   gt  gt  gt  AddMonths datetime datetime 2010 8 25  0  datetime date 2010  8  25   gt  gt  gt  AddMonths datetime datetime 2010 8 25  -12  datetime date 2009  8  25   gt  gt  gt  AddMonths datetime datetime 2010 8 25  -8  datetime date 2009  12  25   gt  gt  gt  AddMonths datetime datetime 2010 8 25  -7  datetime date 2010  1  25  gt  gt  gt

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Another solution  calculate sum of days in month for next n month and add result to current date   import calendar import datetime  def date from now months       today   datetime datetime today        month   today month     year   today year     sum days   0      for i in range int months            month    1          if month    13              month   1             year    1          sum days    calendar monthrange year  month  1       return datetime date today     datetime timedelta sum days   print date from now 12     if to day is 2017-01-01  output  2019-01-01

[python] How do I calculate the date six months from the current date using the datetime Python module?

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