R Lubridate Returns Unwanted Century When Given Two Digit Year

Question

In R, I have a vector of strings representing dates in two different formats:

1. "month/day/year"
2. "month day, year"

The first format has a two digit year so my vector looks something like this:

c("3/18/75", "March 10, 1994", "10/1/80", "June 15, 1979",...)

I want to put the dates in the vector in a standard format. This should be easy with the mdy function from the lubridate package, except when I pass it the first format, it returns an unwanted century.

mdy("3/18/75") returns "2075-03-18 UTC"

Does anyone know how it can return the date in the 20th century? That is "1975-03-18 UTC". Any other solution of how to standardize the dates will be greatly appreciated as well.

I am running version lubridate_1.3.3 if that matters.

Answer 1

You could do it like this:

some_dates <- c("3/18/75", "March 10, 1994", "10/1/80", "June 15, 1979")
dates <- mdy(some_dates)
future_dates <- year(dates) > year(Sys.Date())
year(dates[future_dates]) <- year(dates[future_dates]) - 100

Maybe a better approach would be to remove the ambiguity from your date strings though -- otherwise your code will be wrong when 2075 rolls around ;)

library(stringr)
some_dates <- c('3/18/75', '01/09/53')
str_replace(some_dates, '[0-9]+$', '19\\0')

Or if the two date strings are mixed:

some_dates <- c("3/18/75", "March 10, 1994", "10/1/80", "June 15, 1979")
str_replace(some_dates, '/([0-9]{2}$)', '/19\\1')

Answer 2

lubridate v1.7.4 does. Looking at a 2068 as we speak