If "adding x years to a date" means "increase the YYYY part of a date by
x", then it should be easiest to manipulate the character representation
of your date.
dates <- as.Date(c("2001-01-12","2001-02-12","2001-03-12"))
addYear <- function(d,addyears) {
Y <- as.numeric(strftime(d, "%Y")) + addyears
as.Date(paste(Y, strftime(d,"-%m-%d"), sep = ""))
}
## (There are leapyears with more than 365 days.)
dates + 365*15
addYear(dates,15)
That said, you cannot "go one year ahead" from February 29, unless you
go ahead by 4, 8, 12, ... years (unless the new year is divisible by 100
but not by 400). One possibility would be to leave such dates as Feb 28.
addYear <- function(d,addyears) {
Y <- as.numeric(strftime(d, "%Y")) + addyears
YisLeapyear <- Y%%400==0L | ((Y%%4==0L) & !(Y%%100==0L))
mdpart <- strftime(d,"-%m-%d")
mdpart <- ifelse(mdpart == "-02-29" & YisLeapyear,
mdpart, "-02-28")
as.Date(paste(Y, mdpart, sep = ""))
}
dates <- as.Date(c("2001-01-12","2001-02-12","2001-03-12",
"1899-02-28","1896-02-29","2000-03-01"))
addYear(dates,4)
addYear(dates,5)
addYear(dates,8)
Regards,
Enrico
Am 13.12.2011 04:17, schrieb R. Michael Weylandt:
It depends how your dates are stored, but generally you can just add
365*15 to them. E.g.,
print(x<- Sys.Date())
print(x + 365*15)
So for you,
dataset$dates<- dataset$dates + 365*15
Michael
On Mon, Dec 12, 2011 at 9:39 PM, Ana<rrasterr at gmail.com> wrote:
How do I sum 15 years to my dataset?
original dataset
dates val
1 2001-01-12 1.2
2 2001-02-12 1.2
3 2001-03-12 1.2
result
dates val
1 2016-01-12 1.2
2 2016-02-12 1.2
3 2016-03-12 1.2