How to subtract months from a date in R?

I'm trying to subtract n months from a date as follows:

maturity <- as.Date("2012/12/31")

m <- as.POSIXlt(maturity)

m$mon <- m$mon - 6

but the resulting date is 01-Jul-2012, and not 30-Jun-2012, as I should expect. Is there any short way to get such result?

Thanks in advance

1. seq.Date. Note that June has only 30 days so it cannot give June 31st thus instead it gives July 1st.

   seq(as.Date("2012/12/31"), length = 2, by = "-6 months")[2]

   [1] "2012-07-01"

If we knew it was at month end we could do this:

seq(as.Date(cut(as.Date("2012/12/31"), "month")), length=2, by="-5 month")[2]-1
## "2012-06-30"

2) yearmon. Also if we knew it was month end then we could use the "yearmon" class of the zoo package like this:

library(zoo)
as.Date(as.yearmon(as.Date("2012/12/31")) -.5, frac = 1)
## [1] "2012-06-30"

This converts the date to "yearmon" subtracts 6 months (.5 of a year) and then converts it back to "Date" using frac=1 which means the end of the month (frac=0 would mean the beginning of the month). This also has the advantage over the previous solution that it is vectorized automatically, i.e. as.Date(...) could have been a vector of dates.

Note that if "Date" class is only being used as a way of representing months then we can get rid of it altogether and directly use "yearmon" since that models what we want in the first place:

as.yearmon("2012-12") - .5
## [1] "Jun 2012"

3) mondate. A third solution is the mondate package which has the advantage here that it returns the end of the month 6 months ago without having to know that we are month end:

library(mondate)
mondate("2011/12/31") - 6
## mondate: timeunits="months"
## [1] 2011/06/30

This is also vectorized.

4. lubridate. This lubridate answer has been changed in line with changes in the package:

   library(lubridate) as.Date("2012/12/31") %m-% months(6)

   [1] "2012-06-30"

lubridate is also vectorized.

5. sqldf/SQLite

   library(sqldf) sqldf("select date('2012-12-31', '-6 months') as date")

   date

   1 2012-07-01

or if we knew we were at month end:

sqldf("select date('2012-12-31', '+1 day', '-6 months', '-1 day') as date")
##         date
## 1 2012-06-30

you can use lubridate package for this

library(lubridate)
maturity <- maturity %m-% months(6)

there is no reason for changing the day field.

you can set your day field back to the last day in that month by

day(maturity) <- days_in_month(maturity)

lubridate works correctly with such calculations:

library(lubridate)
as.Date("2000-01-01") - days(1)    # 1999-12-31
as.Date("2000-03-31") - months(1)  # 2000-02-29

but sometimes fails:

as.Date("2000-02-29") - years(1)   # NA, should be 1999-02-28

Technically you cannot add/subtract 1 month to all dates (although you can add/subtract 30 days to all dates, but I suppose, that's not something you want). I think this is what you are looking for

> lubridate::ceiling_date(as.Date("2020-01-31"), unit = "month")
[1] "2020-02-01"
> lubridate::floor_date(as.Date("2020-01-31"), unit = "month")
[1] "2020-01-01"