How to parse an invalid date with lubridate?

Question

I need to parse dates and have a cases like "31/02/2018":

library(lubridate)
> dmy("31/02/2018", quiet = T)
[1] NA

This makes sense as the 31st of Feb does not exist. Is there a way to parse the string "31/02/2018" to e.g. 2018-02-28 ? So not to get an NA, but an actual date?

Thanks.

No, there isn't (because it is not an actual date). You can subset and replace it. — Roland
@RuiBarradas blame the one, who filled the survey... or the one who did not set up a format filter. your choice. ;) — c0bra

Ronak Shah Ronak Shah · Accepted Answer · 2018-11-09T13:05:15

We can write a function assuming you would only have dates which could be higher than the actual date and would have the same format always.

library(lubridate)

get_correct_date <- function(example_date) {
  #Split vector on "/" and get 3 components (date, month, year)
  vecs <- as.numeric(strsplit(example_date, "\\/")[[1]])

  #Check number of days in that month
  last_day_of_month <-  days_in_month(vecs[2])

  #If the input date is higher than actual number of days in that month
  #replace it with last day of that month
  if (vecs[1] > last_day_of_month)
    vecs[1] <- last_day_of_month

  #Paste the date components together to get new modified date
  dmy(paste0(vecs, collapse = "/"))
}


get_correct_date("31/02/2018")
#[1] "2018-02-28"

get_correct_date("31/04/2018")
#[1] "2018-04-30"

get_correct_date("31/05/2018")
#[1] "2018-05-31"

With small modification you can adjust the dates if they have different format or even if some dates are smaller than the first date.

How to parse an invalid date with lubridate?

1 Answers