d <-c("1948/05/28","1952/03/27","1994/12/09")
as.Date(d)
[1] "1948-05-28" "1952-03-27" "1994-12-09"
as.Date("2014-03-20")
[1] "2014-03-20"
as.Date("2014-03-20")-as.Date(d)
Time differences in days
[1] 24037 22638  7041
(as.Date("2014-03-20")-as.Date(d))/365
Time differences in days
[1] 65.85479 62.02192 19.29041

I use (as.Date("2014-03-20")-as.Date(d))/365 to get the years between them ,is there more formal way to do that?

有帮助吗?

解决方案

Try this to ensure that it picks days as the unit. Also 365.25 is slightly better than 365 due to leap years:

as.numeric(difftime(as.Date("2014-03-20"), as.Date(d), unit = "days")) / 365.25

其他提示

lubridate::as.interval() addresses exactly that.

This is a more efficient modification of G. Grothendieck's answer:

(as.integer(as.Date("2014-03-20")) - as.integer(as.Date(d))) / 365.25
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