将正确的世纪添加到日期中，年份提供为“没有世纪的年份”，%y

Question

I have an file with birthdays in %d%b%y format. Some eg.

# "01DEC71" "01AUG54" "01APR81" "01MAY81" "01SEP83" "01FEB59"

I tried to reformat the date as

o108$fmtbirth <- format(as.Date(o108$birth, "%d%b%y"), "%Y/%m/%d")

and this is the result

# "1971/12/01" "2054/08/01" "1981/04/01" "1981/05/01" "1983/09/01" "2059/02/01"

These are birthdays and I see 2054. From this page I see that year values between 00 and 68 are coded as 20 for century. Is there a way to toggle this, in my case I want only 00 to 12 to be coded as 20.

Answer 1

1) chron . chron uses 30 by default so this will convert them converting first to Date (since chron can't read those sorts of dates) reformatting to character with two digit years into a format that chron can understand and finally back to Date.

library(chron)
xx <- c("01AUG11", "01AUG12", "01AUG13") # sample data
as.Date(chron(format(as.Date(xx, "%d%b%y"), "%m/%d/%y")))

That gives a cutoff of 30 but we can get a cutoff of 13 using chron's chron.year.expand option:

library(chron)
options(chron.year.expand = 
     function (y, cut.off = 12, century = c(1900, 2000), ...) {
        chron:::year.expand(y, cut.off = cut.off, century = century, ...)
     }
)

and then repeating the original conversion. For example assuming we had run this options statement already we would get the following with our xx :

> as.Date(chron(format(as.Date(xx, "%d%b%y"), "%m/%d/%y")))
[1] "2011-08-01" "2012-08-01" "1913-08-01"

2) Date only . Here is an alternative that does not use chron. You might want to replace "2012-12-31" with Sys.Date() if the idea is that otherwise future dates are really to be set 100 years back:

d <- as.Date(xx, "%d%b%y")
as.Date(ifelse(d > "2012-12-31", format(d, "19%y-%m-%d"), format(d)))

EDIT: added Date only solution.

Answer 2

请参阅相关线程的响应：

format(as.Date("65-05-14", "%y-%m-%d"), "19%y-%m-%d")

Answer 3

o108$fmtbirth <- format(as.Date(o108$birth, "%d%b%y"), "%Y/%m/%d")

o108$fmtbirth <- as.Date(ifelse(o108$fmtbirth > Sys.Date(), 
                                format(o108$fmtbirth, "19%y-%m-%d"),
                                format(o108$fmtbirth)))