Simplest way to extract date from timestamp

Question

Consider the following timestamp

timestamp <- ymd_hms("2011-08-10 14:00:00", tz = "Pacific/Auckland")

> timestamp
[1] "2011-08-10 14:00:00 NZST"

What is the simplest way to get the day part 2011-08-10 from it, and making sure this day is a proper date and not a string?

Using lubridate::day(timestamp) obviously fails here.

I think you're looking for `lubridate::date`. The difference from `as.Date` is that `date` uses the date as it exists in the assigned time zone instead of converting to UTC first (the default in `as.Date`, though it can take a `tz` parameter). — alistaire, Oct 24 '16 at 15:33
`floor_date(timestamp,"day")` yields a POSIXct with the right timezone... — Erwan LE PENNEC, Oct 24 '16 at 19:35
I think you can also use lubridate and as_date now versus as_datetime `mutate(day=as_datetime(timestamp))` and this will work within a pipe — Bill Perry, Dec 12 '17 at 17:14

score 29 · Accepted Answer · edited May 28 '18 at 13:37

29

This would probably be the simplest way:

date(timestamp)

It will return a date class and not a string.

edited May 28 '18 at 13:37

Stephen Rauch

answered May 28 '18 at 13:11

doron

Praveen Kumar · Answer 2 · 2019-06-27T07:40:31.280

14

Use date instead of day

lubridate::date(timestamp)

edited Jun 27 '19 at 07:40

answered Nov 23 '18 at 07:05

Praveen Kumar

score 4 · Answer 3 · answered Sep 11 '20 at 16:17

4

There is also data.tables as.IDate() function now:

timestamp <- "2011-08-10 14:00:00"
data.table::as.IDate(timestamp)

answered Sep 11 '20 at 16:17

andschar

3 Answers3