Add a month to a Date

Question

I am trying to add a month to a date i have  But then its not possible in a straight manner so far  Following is what i tried   d  lt - as Date  2004-01-31   d   60    1   2004-03-31    Adding wont help as the month wont be overlapped   seq as Date  2004-01-31    by    month   length   2      1   2004-01-31   2004-03-02    Above might work   but again its not straight forward  Also its also adding 30 days or something to the date which has issues like the below  seq as Date  2004-01-31    by    month   length   10       1   2004-01-31   2004-03-02   2004-03-31   2004-05-01   2004-05-31   2004-07-01   2004-07-31   2004-08-31   2004-10-01   2004-10-31    In the above   for the first 2 dates   month haven   t changed   Also the following approach also failed for month but was success for year  d  lt - as POSIXlt as Date  2010-01-01    d year  lt - d year  1 d    1   2011-01-01 UTC  d  lt - as POSIXlt as Date  2010-01-01    d month  lt - d month  1 d      Error in format POSIXlt x  usetz   TRUE    invalid  x  argument   What is the right method to do this

User · Answer

The simplest way is to convert Date to POSIXlt format. Then perform the arithmetic operation as follows:

date_1m_fwd     <- as.POSIXlt("2010-01-01")
date_1m_fwd$mon <- date_1m_fwd$mon +1

Moreover, incase you want to deal with Date columns in data.table, unfortunately, POSIXlt format is not supported.

Still you can perform the add month using basic R codes as follows:

library(data.table)  
dt <- as.data.table(seq(as.Date("2010-01-01"), length.out=5, by="month"))  
dt[,shifted_month:=tail(seq(V1[1], length.out=length(V1)+3, by="month"),length(V1))]

Hope it helps.

User · Answer

It is ambiguous when you say  add a month to a date    Do you mean   add 30 days  increase the month part of the date by 1    In both cases a whole package for a simple addition seems a bit exaggerated   For the first point  of course  the simple   operator will do   d as Date  2010-01-01    d   30    1   2010-01-31    As for the second I would just create a one line function  as simple as that  and with a more general scope    add months  function date n  seq date  by   paste  n   months    length   2  2    You can use it with arbitrary months  including negative    add months d  3    1   2010-04-01  add months d  -3    1   2009-10-01    Of course  if you want to add only and often a single month   add month function date  add months date 1  add month d    1   2010-02-01    If you add one month to 31 of January  since  31th February is meaningless  the best to get the job done is to add the missing 3 days to the following month  March  So correctly   add month as Date  2010-01-31      1   2010-03-03    In case  for some very special reason  you need to put a ceiling to the last available day of the  month  it s a bit longer    add months ceil function  date  n       no ceiling   nC add months date  n      ceiling   day date  01   C add months date  n 1 -1     use ceiling in case of overlapping   if nC gt C  return C    return nC      As usual you could add a single month version   add month ceil function date  add months ceil date 1        So     d as Date  2010-01-31     add month ceil d      1   2010-02-28    d as Date  2010-01-21     add month ceil d      1   2010-02-21    And with decrements     d as Date  2010-03-31     add months ceil d  -1      1   2010-02-28    d as Date  2010-03-21     add months ceil d  -1      1   2010-02-21    Besides you didn t tell if you were interested to a scalar or vector solution  As for the latter     add months v  function date n  as Date sapply date  add months  n   origin  1970-01-01     Note   apply family destroys the class data  that s why it has to be rebuilt   The vector version brings   d c as Date  2010 01 01    as Date  2010 01 31    add months v d 1   1   2010-02-01   2010-03-03    Hope you liked it

User · Answer

Function  m   from lubridate adds one month without exceeding last day of the new month   library lubridate   d  lt - ymd  2012-01-31     1 parsed with  Y- m- d  1   2012-01-31 UTC  d  m   months 1   1   2012-02-29 UTC

User · Answer

Vanilla R has a naive difftime class  but the Lubridate CRAN package lets you do what you ask  require lubridate  d  lt - ymd as Date  2004-01-01     m   months 1  d  1   quot 2004-02-01 quot   Hope that helps

User · Answer

addedMonth  lt - seq as Date  2004-01-01    length 2  by  1 month   2  addedQuarter  lt - seq as Date  2004-01-01    length 2  by  1 quarter   2

User · Answer

I turned antonio s thoughts into a specific function   library DescTools    gt  AddMonths as Date  2004-01-01    1   1   2004-02-01    gt  AddMonths as Date  2004-01-31    1   1   2004-02-29    gt  AddMonths as Date  2004-03-30    -1   1   2004-02-29

User · Answer

Here s a function that doesn t require any packages to be installed  You give it a Date object  or a character that it can convert into a Date   and it adds n months to that date without changing the day of the month  unless the month you land on doesn t have enough days in it  in which case it defaults to the last day of the returned month   Just in case it doesn t make sense reading it  there are some examples below   Function definition  addMonth  lt - function date  n   1     if  n    0  return date     if  n    1    0  stop  Input Error  argument  n  must be an integer          Check to make sure we have a standard Date format   if  class date      character   date   as Date date        Turn the year  month  and day into numbers so we can play with them   y   as numeric substr as character date  1 4     m   as numeric substr as character date  6 7     d   as numeric substr as character date  9 10        Run through the computation   i   0     Adding months   if  n  gt  0       while  i  lt  n         m   m   1       if  m    13           m   1         y   y   1               i   i   1               Subtracting months   else if  n  lt  0       while  i  gt  n         m   m - 1       if  m    0           m   12         y   y - 1               i   i - 1                If past 28th day in base month  make adjustments for February   if  d  gt  28  amp  m    2           If it s a leap year  return the 29th day       if   y    4    0  amp  y    100    0    y    400    0  d   29          Otherwise  return the 28th day       else d   28            If 31st day in base month but only 30 days in end month  return 30th day   else if  d    31  if  m  in  c 1  3  5  7  8  10  12     FALSE  d   30        Turn year  month  and day into strings and put them together to make a Date   y   as character y       If month is single digit  add a leading 0  otherwise leave it alone   if  m  lt  10  m   paste  0   as character m   sep          else m   as character m        If day is single digit  add a leading 0  otherwise leave it alone   if  d  lt  10  d   paste  0   as character d   sep          else d   as character d        Put them together and convert return the result as a Date   return as Date paste y  -  m  -  d  sep             Some examples  Adding months   gt  addMonth  2014-01-31   n   1   1   2014-02-28     February  non-leap year  gt  addMonth  2014-01-31   n   5   1   2014-06-30     June only has 30 days  so day of month dropped to 30  gt  addMonth  2014-01-31   n   24   1   2016-01-31     Increments years when n is a multiple of 12   gt  addMonth  2014-01-31   n   25   1   2016-02-29     February  leap year   Subtracting months   gt  addMonth  2014-01-31   n   -1   1   2013-12-31   gt  addMonth  2014-01-31   n   -7   1   2013-06-30   gt  addMonth  2014-01-31   n   -12   1   2013-01-31   gt  addMonth  2014-01-31   n   -23   1   2012-02-29

User · Answer

mondate  is somewhat similar to  Date  except that adding n adds n months rather than n days    gt  library mondate   gt  d  lt - as Date  2004-01-31    gt  as mondate d    1 mondate  timeunits  months   1  2004-02-29

[r] Add a month to a Date

Examples related to r

Examples related to date

Examples related to date-arithmetic