I am trying to convert a column of dates into Date objects in R, but I can\'t seem to get the desired results. These individuals have birth dates before January 1, 1970, so when
No need for add-on packages, base R is fine. But you need to specify the century:
R> as.Date("1954-01-12")
[1] "1954-01-12"
R>
If you need non-default formats, just specify them:
R> as.Date("19540112", "%Y%m%d")
[1] "1954-01-12"
R>
Edit: In case your data really comes in using the %y%
format, and you happen to make the policy decision that the 19th century is needed
, here is one base R way of doing it:
R> d <- as.Date("540112", "%y%m%d")
R> dlt <- as.POSIXlt(d)
R> dlt$year <- dlt$year - 100
R> as.Date(dlt)
[1] "1954-01-12"
R>
If everything is in the 1900s, its a one-liner - just format it with a two-digit year at the start and slap a 19 on the front and convert to a date. Again. Man this would look cool some %>% stuff:
s = c("1/12/54","1/12/74")
as.Date(format(as.Date(s,format="%d/%m/%y"), "19%y%m%d"), "%Y%m%d")
# [1] "1954-12-01" "1974-12-01"
If years from "69" to "99" are 1800s, then here's another one-liner:
library(dplyr) # for pipe operator:
s %>% as.Date(format="%d/%m/%y") %>%
format("%y%m%d") %>%
(function(d){
paste0(ifelse(d>700101,"18","19"),d)
}) %>%
as.Date("%Y%m%d")
## [1] "1954-12-01" "1874-12-01"
Note not thoroughly tested so might be some off-by-one errors or I've mixed months and days because you need to be ISO8601 Compliant
I would do:
library(lubridate)
x <- as.Date("1/12/54", format = "%m/%d/%y")
year(x) <- 1900 + year(x) %% 100
> x
[1] "1954-01-12"