Convert dd/mm/yy and dd/mm/yyyy to Dates

Question

I have some a character vector with dates in various formats like this

dates <- c(\"23/11/12\", \"20/10/2012\", \"22/10/2012\" ,\"23/11/12\")

Answer 1

You can choose the format based upon input length of date.

y <- ifelse(nchar(dates) == 8, "y", "Y")
as.Date(dates, format = paste0("%d/%m/%", y))

Answer 2

You can use strsplit and nchar to get a subvector of dates where the year is two characters long:

> dates[sapply(strsplit(dates,"/"),function(x)nchar(x)[3]==2)]
[1] "23/11/12" "23/11/12"

Answer 3

You can use parse_date_time from lubridate:

some.dates <- c("23/11/12", "20/10/2012", "22/10/2012" ,"23/11/12")
parse_date_time(some.dates,c('dmy'))
[1] "2012-11-23 UTC" "2012-10-20 UTC" "2012-10-22 UTC" "2012-11-23 UTC"

But , Note that the order of format is important :

some.dates <- c("20/10/2012","23/11/12",  "22/10/2012" ,"23/11/12")
parse_date_time(some.dates,c('dmY','dmy'))

[1] "2012-10-20 UTC" "2012-11-23 UTC" "2012-10-22 UTC" "2012-11-23 UTC"

EDIT

Internally parse_date_time is using guess_formats (which I guess uses some regular expressions):

guess_formats(some.dates,c('dmy'))
       dmy        dmy        dmy        dmy 
"%d/%m/%Y" "%d/%m/%y" "%d/%m/%Y" "%d/%m/%y" 

As mentioned in the comment you can use parse_date_time like this:

as.Date(dates, format = guess_formats(dates,c('dmy')))

Answer 4

If you really wanted to do it in regexp you should have used $ to signify that there was nothing (i.e. end of string) after the last two-digits numbers:

dates[grep('[0-9]{2}/[0-9]{2}/[0-9]{2}$', dates)]
[1] "23/11/12" "23/11/12"

Otherwise, in addition to the other answers you can have a look here and here for other ways of handling multiple date formats.

Answer 5

Following your original attempt at regex based solutions, you may try gsub using this regexp, then converting to any date-time format you wish...

#  Replace 4 digit years with two digit years
short <- gsub( "([0-9]{2})([0-9]{2})$" , "\\2" , dates )
#[1] "23/11/12" "20/10/12" "22/10/12" "23/11/12"


as.Date( short , format = "%d/%m/%y" )
#[1] "2012-11-23" "2012-10-20" "2012-10-22" "2012-11-23"

Answer 6

Here's a base R way for the more general case not (yet) addressed in the unaccepted answers.

dates <- c("23-Jul-2013", "23/11/12", "20/10/2012", "22/10/2012" ,"23/11/12")
fmts <- list('%d-%b-%Y', '%d/%m/%y', '%d/%m/%Y')
d <- mapply(as.Date, list(dates), fmts, SIMPLIFY=FALSE)
max.d <- do.call(function(...) pmax(..., na.rm=TRUE), d)
min.d <- do.call(function(...) pmin(..., na.rm=TRUE), d)
max.d[max.d > Sys.Date()] <- min.d[max.d > Sys.Date()]
max.d
# [1] "2012-11-23" "2012-10-20" "2012-10-22" "2012-11-23"