Return datetime object of previous month

Question

If only timedelta had a month argument in it s constructor   So what s the simplest way to do this   EDIT  I wasn t thinking too hard about this as was pointed out below   Really what I wanted was any day in the last month because eventually I m going to grab the year and month only   So given a datetime object  what s the simplest way to return any datetime object that falls in the previous month

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I use this for government fiscal years where Q4 starts October 1st   Note I convert the date into quarters and undo it as well   import pandas as pd  df  Date      1 1 2020  df  Date     pd to datetime df  Date                  returns 2020-01-01 df  NewDate     df Date - pd DateOffset months 3      returns 2019-10-01  lt ---- answer    For fun  change it to FY Quarter  2019Q4  df  NewDate     df  NewDate   dt year astype str     Q    df  NewDate   dt quarter astype str     Convert  2019Q4  back to 2019-10-01 df  NewDate     pd to datetime df NewDate

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One liner    previous month date    current date - datetime timedelta days current date day 1   replace day current date day

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You could do it in two lines like this   now   datetime now   last month   datetime now year  now month - 1  now day    remember the imports  from datetime import datetime

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For most use cases  what about from datetime import date  current date  date today   current month   current date month last month   current month - 1 if current month    1 else 12   today a month ago   date current date year  last month  current date day   That seems the simplest to me  Note  I ve added the second to last line so that it would work if the current month is January as per  Nick s comment Note 2  In most cases  if the current date is the 31st of a given month the result will be an invalid date as the previous month would not have 31 days  Except for July  amp  August   as noted by  OneCricketeer

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A vectorized  pandas solution is very simple   df  date   - pd DateOffset months 1

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def month sub year  month  sub month       result month   0     result year   0     if month  gt   sub month   12           result month   month -  sub month   12          result year   year -  sub month   12      else          result month   12 -  sub month   12    month         result year   year -  sub month   12   1      return  result year  result month    gt  gt  gt  month sub 2015  7  1       2015  6   gt  gt  gt  month sub 2015  7  -1   2015  8   gt  gt  gt  month sub 2015  7  13   2014  6   gt  gt  gt  month sub 2015  7  -14   2016  9

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I think the simple way is to use DateOffset from Pandas like so  import pandas as pd date 1   pd to datetime  quot 2013-03-31 quot   format  quot  Y- m- d quot   - pd DateOffset months 1   The result will be a Timestamp object

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Try this  def monthdelta date  delta       m  y    date month delta    12  date year     date month  delta-1     12     if not m  m   12     d   min date day   31          29 if y 4  0 and  not y 100  0 or y 400    0  else 28          31 30 31 30 31 31 30 31 30 31  m-1       return date replace day d month m  year y    gt  gt  gt  for m in range -12  12       print monthdelta datetime now    m         2009-08-06 16 12 27 823000 2009-09-06 16 12 27 855000 2009-10-06 16 12 27 870000 2009-11-06 16 12 27 870000 2009-12-06 16 12 27 870000 2010-01-06 16 12 27 870000 2010-02-06 16 12 27 870000 2010-03-06 16 12 27 886000 2010-04-06 16 12 27 886000 2010-05-06 16 12 27 886000 2010-06-06 16 12 27 886000 2010-07-06 16 12 27 886000 2010-08-06 16 12 27 901000 2010-09-06 16 12 27 901000 2010-10-06 16 12 27 901000 2010-11-06 16 12 27 901000 2010-12-06 16 12 27 901000 2011-01-06 16 12 27 917000 2011-02-06 16 12 27 917000 2011-03-06 16 12 27 917000 2011-04-06 16 12 27 917000 2011-05-06 16 12 27 917000 2011-06-06 16 12 27 933000 2011-07-06 16 12 27 933000  gt  gt  gt  monthdelta datetime 2010 3 30   -1  datetime datetime 2010  2  28  0  0   gt  gt  gt  monthdelta datetime 2008 3 30   -1  datetime datetime 2008  2  29  0  0   Edit Corrected to handle the day as well  Edit See also the answer from puzzlement which points out a simpler calculation for d  d   min date day  calendar monthrange y  m  1

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Returns last day of last month   gt  gt  gt  import datetime  gt  gt  gt  datetime datetime now   - datetime timedelta days datetime datetime now   day  datetime datetime 2020  9  30  14  13  15  67582   Returns the same day last month   gt  gt  gt  x   datetime datetime now   - datetime timedelta days datetime datetime now   day   gt  gt  gt  x replace day datetime datetime now   day  datetime datetime 2020  9  7  14  22  14  362421

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Here is some code to do just that  Haven t tried it out myself     def add one month t          Return a  datetime date  or  datetime datetime   as given  that is     one month earlier       Note that the resultant day of the month might change if the following     month has fewer days            gt  gt  gt  add one month datetime date 2010  1  31           datetime date 2010  2  28              import datetime     one day   datetime timedelta days 1      one month later   t   one day     while one month later month    t month     advance to start of next month         one month later    one day     target month   one month later month     while one month later day  lt  t day     advance to appropriate day         one month later    one day         if one month later month    target month     gone too far             one month later -  one day             break     return one month later  def subtract one month t          Return a  datetime date  or  datetime datetime   as given  that is     one month later       Note that the resultant day of the month might change if the following     month has fewer days            gt  gt  gt  subtract one month datetime date 2010  3  31           datetime date 2010  2  28              import datetime     one day   datetime timedelta days 1      one month earlier   t - one day     while one month earlier month    t month or one month earlier day  gt  t day          one month earlier -  one day     return one month earlier

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If only timedelta had a month argument   in it s constructor  So what s the   simplest way to do this    What do you want the result to be when you subtract a month from  say  a date that is March 30   That is the problem with adding or subtracting months  months have different lengths   In some application an exception is appropriate in such cases  in others  the last day of the previous month  is OK to use  but that s truly crazy arithmetic  when subtracting a month then adding a month is not overall a no-operation    in others yet you ll want to keep in addition to the date some indication about the fact  e g    I m saying Feb 28 but I really would want Feb 30 if it existed   so that adding or subtracting another month to that can set things right again  and the latter obviously requires a custom class holding a data plus s thing else    There can be no real solution that is tolerable for all applications  and you have not told us what your specific app s needs are for the semantics of this wretched operation  so there s not much more help that we can provide here

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import datetime date str    08 01 2018  format str     d  m  Y  datetime obj   datetime datetime strptime date str  format str     datetime obj replace month datetime obj month-1    Simple solution  no need for special libraries

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You can use below given function to get date before after X month    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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Simplest Way that i have tried Just now   from datetime import datetime from django utils import timezone      current   timezone now   if current month    1       month   12 else       month   current month - 1 current   datetime current year  month  current day

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I Used the following code to go back n Months from a specific Date   your date    datetime strptime input date    Y- m- d     to convert date 2016-01-01  to timestamp start date your date     start from current date   Calculate Month for i in range 0 n       n   number of months you need to go back     start date start date replace day 1      1st day of current month     start date start date-timedelta days 1   last day of previous month   Calculate Day if start date day gt your date day          start date start date replace day your date day               print start date   For eg  input date   28 12 2015 Calculate 6 months previous date   I  CALCULATE MONTH   This step will give you the start date as 30 06 2015  Note that after the calculate month step you will get the last day of the required month   II CALCULATE DAY  Condition if start date day your date day   checks whether the day from input date is present in the required month  This handles condition where input date is 31 or 30  and the required month has less than 31 or 30 in case of feb  days  It handles leap year case as well For Feb   After this step you will get result as 28 06 2015  If this condition is not satisfied  the start date remains the last date of the previous month  So if you give 31 12 2015 as input date and want 6 months previous date  it will give you 30 06 2015

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If all you want is any day in the last month  the simplest thing you can do is subtract the number of days from the current date  which will give you the last day of the previous month   For instance  starting with any date    gt  gt  gt  import datetime                                                                                                                                                                   gt  gt  gt  today   datetime date today                                                                                                                                                       gt  gt  gt  today datetime date 2016  5  24    Subtracting the days of the current date we get    gt  gt  gt  last day previous month   today - datetime timedelta days today day   gt  gt  gt  last day previous month datetime date 2016  4  30    This is enough for your simplified need of any day in the last month   But now that you have it  you can also get any day in the month  including the same day you started with  i e  more or less the same as subtracting a month     gt  gt  gt  same day last month   last day previous month replace day today day   gt  gt  gt  same day last month datetime date 2016  4  24    Of course  you need to be careful with 31st on a 30 day month or the days missing from February  and take care of leap years   but that s also easy to do    gt  gt  gt  a date   datetime date 2016  3  31                                                                                                                                                gt  gt  gt  last day previous month   a date - datetime timedelta days a date day   gt  gt  gt  a date minus month             last day previous month replace day a date day          if a date day  lt  last day previous month day         else last day previous month        gt  gt  gt  a date minus month datetime date 2016  2  29

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I think this answer is quite readable   def month delta dt  delta       year delta  month   divmod dt month   delta  12       if month    0            convert a 0 to december         month   12         if delta  lt  0                if moving backwards  then it s december of last year             year delta -  1      year   dt year   year delta      return dt replace month month  year year   for delta in range -20  21       print delta   - gt    month delta datetime 2011  1  1   delta    -20 - gt  2009-05-01 00 00 00 -19 - gt  2009-06-01 00 00 00 -18 - gt  2009-07-01 00 00 00 -17 - gt  2009-08-01 00 00 00 -16 - gt  2009-09-01 00 00 00 -15 - gt  2009-10-01 00 00 00 -14 - gt  2009-11-01 00 00 00 -13 - gt  2009-12-01 00 00 00 -12 - gt  2010-01-01 00 00 00 -11 - gt  2010-02-01 00 00 00 -10 - gt  2010-03-01 00 00 00 -9 - gt  2010-04-01 00 00 00 -8 - gt  2010-05-01 00 00 00 -7 - gt  2010-06-01 00 00 00 -6 - gt  2010-07-01 00 00 00 -5 - gt  2010-08-01 00 00 00 -4 - gt  2010-09-01 00 00 00 -3 - gt  2010-10-01 00 00 00 -2 - gt  2010-11-01 00 00 00 -1 - gt  2010-12-01 00 00 00 0 - gt  2011-01-01 00 00 00 1 - gt  2011-02-01 00 00 00 2 - gt  2011-03-01 00 00 00 3 - gt  2011-04-01 00 00 00 4 - gt  2011-05-01 00 00 00 5 - gt  2011-06-01 00 00 00 6 - gt  2011-07-01 00 00 00 7 - gt  2011-08-01 00 00 00 8 - gt  2011-09-01 00 00 00 9 - gt  2011-10-01 00 00 00 10 - gt  2011-11-01 00 00 00 11 - gt  2012-12-01 00 00 00 12 - gt  2012-01-01 00 00 00 13 - gt  2012-02-01 00 00 00 14 - gt  2012-03-01 00 00 00 15 - gt  2012-04-01 00 00 00 16 - gt  2012-05-01 00 00 00 17 - gt  2012-06-01 00 00 00 18 - gt  2012-07-01 00 00 00 19 - gt  2012-08-01 00 00 00 20 - gt  2012-09-01 00 00 00

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A variation on Duncan s answer  I don t have sufficient reputation to comment   which uses calendar monthrange to dramatically simplify the computation of the last day of the month   import calendar def monthdelta date  delta       m  y    date month delta    12  date year     date month  delta-1     12     if not m  m   12     d   min date day  calendar monthrange y  m  1       return date replace day d month m  year y    Info on monthrange from Get Last Day of the Month in Python

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After the original question s edit to  any datetime object in the previous month   you can do it pretty easily by subtracting 1 day from the first of the month   from datetime import datetime  timedelta  def a day in previous month dt      return dt replace day 1  - timedelta days 1

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Some time ago I came across the following algorithm which works very well for incrementing and decrementing months on either a date or datetime  CAVEAT  This will fail if day is not available in the new month  I use this on date objects where day    1 always  Python 3 x  def increment month d  add 1       return date d year  d month add-1   12   d month add-1    12 1  1   For Python 2 7 change the   12 to just  12 since integer division is implied  I recently used this in a defaults file when a script started to get these useful globals  MONTH THIS   datetime date today   MONTH THIS   datetime date MONTH THIS year  MONTH THIS month  1   MONTH 1AGO   datetime date MONTH THIS year  MONTH THIS month-2   12                              MONTH THIS month-2    12 1  1   MONTH 2AGO   datetime date MONTH THIS year  MONTH THIS month-3   12                              MONTH THIS month-3    12 1  1

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You can use the third party dateutil module  PyPI entry here    import datetime import dateutil relativedelta  d   datetime datetime strptime  2013-03-31     Y- m- d   d2   d - dateutil relativedelta relativedelta months 1  print d2   output   2013-02-28 00 00 00

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Given a  year month  tuple  where month goes from 1-12  try this    gt  gt  gt  from datetime import datetime  gt  gt  gt  today   datetime today    gt  gt  gt  today datetime datetime 2010  8  6  10  15  21  310000   gt  gt  gt  thismonth   today year  today month  gt  gt  gt  thismonth  2010  8   gt  gt  gt  lastmonth   lambda  yr mo     y m 1  for y m in  divmod  yr 12 mo-2   12     0   gt  gt  gt  lastmonth thismonth   2010  7   gt  gt  gt  lastmonth   2010 1     2009  12    Assumes there are 12 months in every year

[python] Return datetime object of previous month

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