Reputation: 34763
Am I missing something? I can't figure out how to convert the following to Date
s, where day of the month (%d
) has the ordinal suffixes -st
, -nd
, -rd
, -th
:
ord_dates <- c("September 1st, 2016", "September 2nd, 2016",
"September 3rd, 2016", "September 4th, 2016")
?strptime
doesn't appear to list a shorthand for the ordinal suffix, and it isn't handled automagically:
as.Date(ord_dates, format = c("%B %d, %Y"))
#[1] NA NA NA NA
Is there a token for handling ignored characters in the format
argument? A token I'm missing?
Best I can come up with is (there may a shorter regex, but same idea):
as.Date(gsub("([0-9]+)(st|nd|rd|th)", "\\1", ord_dates), format = "%B %d, %Y")
# [1] "2016-09-01" "2016-09-02" "2016-09-03" "2016-09-04"
Seems like this sort of data should be relatively common; am I missing something?
Upvotes: 13
Views: 2164
Reputation: 67
I came across this thread after searching for "ordinal dates" in R. I tried using the code suggested above, but found that it removed ALL text following the date. This slight variation searches instead for numbers followed by two lowercase letters.
sub("([0-9])[a-z]{2}", "\\1", ord_dates)
All together, the solution would be
as.Date(sub("([0-9])[a-z]{2}", "\\1", ord_dates), format = "%B %d, %Y")
Upvotes: 0
Reputation: 1751
Enjoy the power of lubridate
:
library(lubridate)
mdy(ord_dates)
[1] "2016-09-01" "2016-09-02" "2016-09-03" "2016-09-04"
Internally, lubridate
doesn't have any special conversion specifications which enable this. Rather, lubridate
first uses (by smart guessing) the format "%B %dst, %Y"
. This gets the first element of ord_dates
.
It then checks for NA
s and repeats its smart guessing on the remaining elements, settling on "%B %dnd, %Y"
to get the second element. It continues in this way until there are no NA
s left (which happens in this case after 4 iterations), or until its smart guessing fails to turn up a likely format candidate.
You can imagine this makes lubridate
slower, and it does -- about half the speed of just using the smart regex suggested by @alistaire above:
set.seed(109123)
ord_dates <- sample(
c("September 1st, 2016", "September 2nd, 2016",
"September 3rd, 2016", "September 4th, 2016"),
1e6, TRUE
)
library(microbenchmark)
microbenchmark(times = 10L,
lubridate = mdy(ord_dates),
base = as.Date(sub("\\D+,", "", ord_dates),
format = "%B %e %Y"))
# Unit: seconds
# expr min lq mean median uq max neval cld
# lubridate 2.167957 2.219463 2.290950 2.252565 2.301725 2.587724 10 b
# base 1.183970 1.224824 1.218642 1.227034 1.228324 1.229095 10 a
The obvious advantage in lubridate
's favor being its conciseness and flexibility.
Upvotes: 11