CODEWITHSUNDEEP
rdatetimelubridate
S.Perera
S.Perera

Reputation: 894

`as_datetime()` in lubridate package

Why there is a strange behavior in the following expressions (why the dates are different):

today("America/New_York")
[1] "2021-02-28"

as_datetime(today("America/New_York"), tz = "America/New_York")
[1] "2021-02-27 19:00:00 EST"

Also, the local time in New York is 19:09. Is it rounding this time to get "19:00:00"?

Upvotes: 0

Views: 1500

Answers (2)

neilfws
neilfws

Reputation: 33772

I think this is a consequence of using today(), which returns a Date not a DateTime. It appears as if applying as_datetime starts with the assumption of UTC and then substracts the time zone difference.

The output of today() is type Date:

str(today(tzone = "America/New_York"))
 Date[1:1], format: "2021-02-28"

Without time zone information, as_datetime defaults to UTC. Without time data, it seems to assume that the time is midnight on the given date:

as_datetime(today(tzone = "America/New_York"))
[1] "2021-02-28 UTC"

str(as_datetime(today(tzone = "America/New_York")))
 POSIXct[1:1], format: "2021-02-28"

So when we add time zone information it subtracts 5 hours from midnight, giving 19:00 on the previous date:

as_datetime(today(tzone = "America/New_York"), tz = "America/New_York")
[1] "2021-02-27 19:00:00 EST"

Note the different methods in as_datetime:

showMethods("as_datetime")
Function: as_datetime (package lubridate)
x="ANY"
x="character"
x="Date"
    (inherited from: x="ANY")
x="numeric"
x="POSIXt"

Upvotes: 1

akrun
akrun

Reputation: 886938

Instead of today, use now

as_datetime(now("America/New_York"), tz = "America/New_York")
#[1] "2021-02-28 19:16:36 EST"

According to ?now

now - the current datetime as a POSIXct object

whereas today - returns the Date which is truncated

str(today())
#Date[1:1], format: "2021-02-28"

Upvotes: 2

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