Lubridate mdy function - r

Question

I'm trying to convert the following and am not successful with one of the dates [1]. "4/2/10" becomes "0010-04-02".
Is there a way to correct this?
thanks,
Vivek
data <- data.frame(initialDiagnose = c("4/2/10","14.01.2009", "9/22/2005",
"4/21/2010", "28.01.2010", "09.01.2009", "3/28/2005",
"04.01.2005", "04.01.2005", "Created on 9/17/2010", "03 01 2010"))
mdy <- mdy(data$initialDiagnose)
dmy <- dmy(data$initialDiagnose)
mdy[is.na(mdy)] <- dmy[is.na(mdy)] # some dates are ambiguous, here we give
data$initialDiagnose <- mdy # mdy precedence over dmy
data
initialDiagnose
1 0010-04-02
2 2009-01-14
3 2005-09-22
4 2010-04-21
5 2010-01-28
6 2009-09-01
7 2005-03-28
8 2005-04-01
9 2005-04-01
10 2010-09-17
11 2010-03-01

Answer 1

I think this is occurring because the mdy() function prefers to match the year with %Y (the actual year) over %y (2 digit abbreviation for the year, defaulting to 19XX or 20XX).
There is a workaround, though. I took a look at the help files for lubridate::parse_date_time (?parse_date_time), and near the bottom of the help file, there is an example for adding an argument that prefers matching with the %y format over the %Y format for the year. The relevant bit of code from the help file:
## ** how to use `select_formats` argument **
## By default %Y has precedence:
parse_date_time(c("27-09-13", "27-09-2013"), "dmy")
## [1] "13-09-27 UTC" "2013-09-27 UTC"
## to give priority to %y format, define your own select_format function:
my_select <- function(trained){
n_fmts <- nchar(gsub("[^%]", "", names(trained))) + grepl("%y", names(trained))*1.5
names(trained[ which.max(n_fmts) ])
}
parse_date_time(c("27-09-13", "27-09-2013"), "dmy", select_formats = my_select)
## '[1] "2013-09-27 UTC" "2013-09-27 UTC"
So, for your example, you can adapt this code and replace the mdy <- mdy(data$initialDiagnose) line with this:
# Define a select function that prefers %y over %Y. This is copied
# directly from the help files
my_select <- function(trained){
n_fmts <- nchar(gsub("[^%]", "", names(trained))) + grepl("%y", names(trained))*1.5
names(trained[ which.max(n_fmts) ])
}
# Parse as mdy dates
mdy <- parse_date_time(data$initialDiagnose, "mdy", select_formats = my_select)
# [1] "2010-04-02 UTC" NA "2005-09-22 UTC" "2010-04-21 UTC" NA
# [6] "2009-09-01 UTC" "2005-03-28 UTC" "2005-04-01 UTC" "2005-04-01 UTC" "2010-09-17 UTC"
#[11] "2010-03-01 UTC"
And running the remaining lines of code from your question, it gives me this data frame as the result:
initialDiagnose
1 2010-04-02
2 2009-01-14
3 2005-09-22
4 2010-04-21
5 2010-01-28
6 2009-09-01
7 2005-03-28
8 2005-04-01
9 2005-04-01
10 2010-09-17
11 2010-03-01