问题:

I have a date in R, e.g.:

dt = as.Date('2010/03/17')

I would like to subtract 2 years from this date, without worrying about leap years and such issues, getting as.Date('2010-03-17').

How would I do that?

回答1:

The easiest thing to do is to convert it into POSIXlt and subtract 2 from the years slot.

> d <- as.POSIXlt(as.Date('2010/03/17'))
> d$year <- d$year-2
> as.Date(d)
[1] "2008-03-17"

See this related question: How to subtract days in R?.

回答2:

With lubridate

library(lubridate)
ymd("2010/03/17") - years(2)

回答3:

You could use seq:

R> dt = as.Date('2010/03/17')
R> seq(dt, length=2, by="-2 years")[2]
[1] "2008-03-17"

回答4:

If leap days are to be taken into account then I'd recommend using this lubridate function to subtract months, as other methods will return either March 1st or NA:

> library(lubridate)
> dt %m-% months(12*2)
[1] "2008-03-17"

# Try with leap day
> leapdt <- as.Date('2016/02/29')
> leapdt %m-% months(12*2)
[1] "2014-02-28"

回答5:

Same answer than the one by rcs but with the possibility to operate it on a vector (to answer to MichaelChirico, I can't comment I don't have enough rep):

R> unlist(lapply(c("2015-12-01", "2016-12-01"), 
      function(x) { return(as.character(seq(as.Date(x), length=2, by="-1 years")[2])) }))
 [1] "2014-12-01" "2015-12-01"