Python timedelta in years

Question

I need to check if some number of years have been since some date  Currently I ve got timedelta from datetime module and I don t know how to convert it to years

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In the end what you have is a maths issue  If every 4 years we have an extra day lets then dived the timedelta in days  not by 365 but 365 4   1  that would give you the amount of 4 years  Then divide it again by 4   timedelta     365 4   1    4   timedelta   4    365 4  1

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Yet another 3rd party lib not mentioned here is mxDateTime  predecessor of both python datetime and 3rd party timeutil  could be used for this task   The aforementioned yearsago would be   from mx DateTime import now  RelativeDateTime  def years ago years  from date None       if from date    None          from date   now       return from date-RelativeDateTime years years    First parameter is expected to be a DateTime instance   To convert ordinary datetime to DateTime you could use this for 1 second precision    def DT from dt s t       return DT DateTimeFromTicks time mktime t timetuple       or this for 1 microsecond precision   def DT from dt u t       return DT DateTime t year  t month  t day  t hour    t minute  t second   t microsecond   1e-6    And yes  adding the dependency for this single task in question would definitely be an overkill compared even with using timeutil  suggested by Rick Copeland

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This is the solution I worked out  I hope can help  -   def menor edad legal birthday           returns true if aged lt 18 in days          try           today   time localtime                                    fa divuit anys date year today tm year-18  month today tm mon  day today tm mday           if birthday gt fa divuit anys              return True         else              return False                  except Exception  ex edad          logging error  Error menor de edad   s    ex edad          return True

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I came across this question and found Adams answer the most helpful https   stackoverflow com a 765862 2964689  But there was no python example of his method but here s what I ended up using   input  datetime object  output  integer age in whole years  def age birthday       birthday   birthday date       today   date today        years   today year - birthday year      if  today month  lt  birthday month or         today month    birthday month and today day  lt  birthday day             years   years - 1      return years

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You need more than a timedelta to tell how many years have passed  you also need to know the beginning  or ending  date    It s a leap year thing      Your best bet is to use the dateutil relativedelta object  but that s a 3rd party module   If you want to know the datetime that was n years from some date  defaulting to right now   you can do the following    from dateutil relativedelta import relativedelta  def yearsago years  from date None       if from date is None          from date   datetime now       return from date - relativedelta years years    If you d rather stick with the standard library  the answer is a little more complex    from datetime import datetime def yearsago years  from date None       if from date is None          from date   datetime now       try          return from date replace year from date year - years      except ValueError            Must be 2 29          assert from date month    2 and from date day    29   can be removed         return from date replace month 2  day 28                                   year from date year-years    If it s 2 29  and 18 years ago there was no 2 29  this function will return 2 28   If you d rather return 3 1  just change the last return statement to read        return from date replace month 3  day 1                               year from date year-years    Your question originally said you wanted to know how many years it s been since some date   Assuming you want an integer number of years  you can guess based on 365 25 days per year and then check using either of the yearsago functions defined above    def num years begin  end None       if end is None          end   datetime now       num years   int  end - begin  days   365 25      if begin  gt  yearsago num years  end           return num years - 1     else          return num years

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I would use datetime date data type instead  as it is simpler when it comes to checking how many years  months and days have passed  now   date today   birthday   date 1993  4  4  print  quot you are quot   now year - birthday year   quot years  quot   now month - birthday month   quot months and quot     now day - birthday day   quot days old quot    Output  you are 27 years  7 months and 11 days old  I use timedelta when I need to perform arithmetic on a specific date  age   now - birthday print  quot addition of days to a date   quot   birthday   timedelta days age days    Output  addition of days to a date   2020-11-15

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Late to the party  but this will give you the age  in years  accurately and easily  b   birthday today   datetime datetime today   age   today year - b year    today month - b month  gt  0 or                                today month    b month  gt  0 and                                today day - b day  gt  0

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First off  at the most detailed level  the problem can t be solved exactly   Years vary in length  and there isn t a clear  right choice  for year length     That said  get the difference in whatever units are  natural   probably seconds  and divide by the ratio between that and years   E g   delta in days    365 25  delta in seconds    365 25 24 60 60       or whatever   Stay away from months  since they are even less well defined than years

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def age dob       import datetime     today   datetime date today        if today month  lt  dob month or          today month    dob month and today day  lt  dob day           return today year - dob year - 1     else          return today year - dob year   gt  gt  gt  import datetime  gt  gt  gt  datetime date today   datetime date 2009  12  1   gt  gt  gt  age datetime date 2008  11  30   1  gt  gt  gt  age datetime date 2008  12  1   1  gt  gt  gt  age datetime date 2008  12  2   0

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this function returns the difference in years between two dates  taken as strings in ISO format  but it can easily modified to take in any format   import time def years earlydateiso   laterdateiso          difference in years between two dates in ISO format         ed    time strptime earlydateiso    Y- m- d       ld    time strptime laterdateiso    Y- m- d        switch dates if needed     if  ld  lt  ed          ld   ed   ed   ld                  res   ld 0  - ed  0      if res  gt  0          if ld 1  lt  ed 1               res -  1         elif  ld 1     ed 1               if ld 2  lt  ed 2                   res -  1     return res

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I liked John Mee s solution for its simplicity  and I am not that concerned about how  on Feb 28 or March 1 when it is not a leap year  to determine age of people born on Feb 29  But here is a tweak of his code which I think addresses the complaints   def age dob       import datetime     today   datetime date today       age   today year - dob year     if   today month    dob month    2 and          today day    28 and dob day    29             pass     elif today month  lt  dob month or          today month    dob month and today day  lt  dob day           age -  1     return age

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How exact do you need it to be   td days   365 25 will get you pretty close  if you re worried about leap years

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Here s a updated DOB function  which calculates birthdays the same way humans do   import datetime import locale     Source  https   en wikipedia org wiki February 29 PRE          US        TW     POST          GB        HK       def get country        code      locale getlocale       try          return code split      1      except IndexError          raise Exception  Country cannot be ascertained from locale      def get leap birthday year       country   get country       if country in PRE          return datetime date year  2  28      elif country in POST          return datetime date year  3  1      else          raise Exception  It is unknown whether your country treats leap year                            birthdays as being on the 28th of February or                            the 1st of March  Please consult your country  s                            legal code for in order to ascertain an answer    def age dob       today   datetime date today       years   today year - dob year      try          birthday   datetime date today year  dob month  dob day      except ValueError as e          if dob month    2 and dob day    29              birthday   get leap birthday today year          else              raise e      if today  lt  birthday          years -  1     return years  print age datetime date 1988  2  29

User · Answer

Get the number of days  then divide by 365 2425  the mean Gregorian year  for years  Divide by 30 436875  the mean Gregorian month  for months

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import datetime  def check if old enough years needed  old date        limit date   datetime date old date year   years needed   old date month  old date day       today   datetime datetime now   date        old enough   False      if limit date  lt   today          old enough   True      return old enough    def test ages         years needed   30      born date Logan   datetime datetime 1988  3  5       if check if old enough years needed  born date Logan           print  Logan is old enough       else          print  Logan is not old enough         born date Jessica   datetime datetime 1997  3  6       if check if old enough years needed  born date Jessica           print  Jessica is old enough       else          print  Jessica is not old enough     test ages     This is the code that the Carrousel operator was running in Logan s Run film     https   en wikipedia org wiki Logan 27s Run  film

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I ll suggest Pyfdate      What is pyfdate       Given Python s goal to be a powerful and easy-to-use scripting   language  its features for working   with dates and times are not as   user-friendly as they should be  The   purpose of pyfdate is to remedy that   situation by providing features for   working with dates and times that are   as powerful and easy-to-use as the   rest of Python    the tutorial

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If you re trying to check if someone is 18 years of age  using timedelta will not work correctly on some edge cases because of leap years   For example  someone born on January 1  2000  will turn 18 exactly 6575 days later on January 1  2018  5 leap years included   but someone born on January 1  2001  will turn 18 exactly 6574 days later on January 1  2019  4 leap years included    Thus  you if someone is exactly 6574 days old  you can t determine if they are 17 or 18 without knowing a little more information about their birthdate   The correct way to do this is to calculate the age directly from the dates  by subtracting the two years  and then subtracting one if the current month day precedes the birth month day

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Even though this thread is already dead  might i suggest a working solution for this very same problem i was facing  Here it is  date is a string in the format dd-mm-yyyy    def validatedate date       parts   date strip   split  -        if len parts     3 and False not in  x isdigit   for x in parts            birth   datetime date int parts 2    int parts 1    int parts 0            today   datetime date today            b    birth year   10000     birth month   100     birth day          t    today year   10000     today month   100     today day           if  t - 18   10000   gt   b              return True      return False

[python] Python timedelta in years

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