Add one year in current date PYTHON

Question

I have fetched a date from database with the following variable      i operation date      with which I got a value like   April 1  2013   I need to add one year to the above  so that I can get   April 1  2014   Please suggest  how can I do this

User · Answer

You can use Python-dateutil s relativedelta to increment a datetime object while remaining sensitive to things like leap years and month lengths   Python-dateutil comes packaged with matplotlib if you already have that   You can do the following   from dateutil relativedelta import relativedelta  new date   old date   relativedelta years 1     This answer was given by  Max to a similar question      But if your date is a string  i e  not already a datetime object  you can convert it using datetime   from datetime import datetime from dateutil relativedelta import relativedelta  your date string    April 1  2012  format string     B  d   Y   datetime object   datetime strptime your date string  format string  date   new date   datetime object   relativedelta years 1  new date string   datetime strftime new date  format string  replace   0          new date string will contain  April 1  2013    NB  Unfortunately  datetime only outputs day values as  decimal numbers  - i e  with leading zeros if they re single digit numbers   The  replace   at the end is a workaround to deal with this issue copied from  Alex Martelli  see this question for his and other approaches to this problem

User · Answer

It seems from your question that you would like to simply increment the year of your given date rather than worry about leap year implications  You can use the date class to do this by accessing its member year    from datetime import date startDate   date 2012  12  21     reconstruct date fully endDate   date startDate year   1  startDate month  startDate day    replace year only endDate   startDate replace startDate year   1    If you re having problems creating one given your format  let us know

User · Answer

Here s one more answer that I ve found to be pretty concise and doesn t use external packages  import datetime as dt import calendar    Today  in  dt date  type day   dt datetime now   date    one year delta   dt timedelta days 366 if   day month  gt   3 and calendar isleap day year 1   or                                              day month  lt  3 and calendar isleap day year    else 365     Add one year to the current date print day   one year delta

User · Answer

This is what I do when I need to add months or years and don t want to import more libraries  Just create a datetime date   object  call add month date  to add a month and add year date  to add a year   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

User · Answer

AGSM s answer shows a convenient way of solving this problem using the python-dateutil package  But what if you don t want to install that package  You could solve the problem in vanilla Python like this   from datetime import date  def add years d  years          Return a date that s  years  years after the date  or datetime      object  d   Return the same calendar date  month and day  in the     destination year  if it exists  otherwise use the following day      thus changing February 29 to March 1                try          return d replace year   d year   years      except ValueError          return d    date d year   years  1  1  - date d year  1  1     If you want the other possibility  changing February 29 to February 28  then the last line should be changed to           return d    date d year   years  3  1  - date d year  3  1

User · Answer

convert it into python datetime object if it isn t already  then add deltatime  one years later   Your date   datetime timedelta days  years days per year      for your case days 365   you can have condition to check if the year is leap or no and adjust days accordingly  you can add as many years as you want

User · Answer

Look at this      usr bin python  import datetime  def addYears date  years       result   date   datetime timedelta 366   years      if years  gt  0          while result year - date year  gt  years or date month  lt  result month or date day  lt  result day              result    datetime timedelta -1      elif years  lt  0          while result year - date year  lt  years or date month  gt  result month or date day  gt  result day              result    datetime timedelta 1      print  input   s output   s     date  result      return result   Example usage   addYears datetime date 2012 1 1   -1  addYears datetime date 2012 1 1   0  addYears datetime date 2012 1 1   1  addYears datetime date 2012 1 1   -10  addYears datetime date 2012 1 1   0  addYears datetime date 2012 1 1   10    And output of this example   input  2012-01-01 output  2011-01-01 input  2012-01-01 output  2012-01-01 input  2012-01-01 output  2013-01-01 input  2012-01-01 output  2002-01-01 input  2012-01-01 output  2012-01-01 input  2012-01-01 output  2022-01-01

User · Answer

Another way would be to use pandas  quot DateOffset quot  class link -https   pandas pydata org pandas-docs stable reference api pandas tseries offsets DateOffset html Using ASGM s code above in the answers   from datetime import datetime import pandas as pd  your date string    quot April 1  2012 quot  format string    quot  B  d   Y quot   datetime object   datetime strptime your date string  format string  date   new date   datetime object   pd DateOffset years 1   new date date    It will return the datetime object with the added year  Something like this - datetime date 2013  4  1

[python] Add one year in current date PYTHON

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