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