
Question:
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?
Solution: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?.
Solution:2
With lubridate
library(lubridate) ymd("2010/03/17") - years(2)
Solution: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"
Solution: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"
Solution: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"
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