Extract time from timestamp?

Question

Essentially, I want only the hour, minute, and seconds from a column of timestamps I have in R, because I want to view how often different data points occur throughout diffe

Answer 1

We can use strptime to convert to a datetime class and then format to extract the hour:min:sec.

dtime <- strptime(str1, "%Y-%m-%dT%H:%M:%SZ")
format(dtime, "%H:%M:%S")
#[1] "17:07:36"

If the OP wants to have the hour, min, sec as separate columns

read.table(text=format(dtime, "%H:%M:%S"), sep=":", header=FALSE)
#  V1 V2 V3
#1 17  7 36

---

Another option is using lubridate

library(lubridate)
format(ymd_hms(str1), "%H:%M:%S")
#[1] "17:07:36"

data

str1 <- "2008-08-07T17:07:36Z"

Answer 2

I think you are expecting this...

Sys.time()
[1] "2016-04-19 11:09:30 IST"
format(Sys.time(),format = '%T')
[1] "11:09:30"

if you want to give your own timestamp, then use bellow code:

format(as.POSIXlt("2016-04-19 11:02:22 IST"),format = '%T')
[1] "11:02:22"

Answer 3

Just

x <- '2008-08-07T17:07:36Z'
substr(x, 12, 19)
#[1] "17:07:36"

...will do it if the timestamp is consistent, which I imagine it would be given it is an ISO_8601 ( https://en.wikipedia.org/wiki/ISO_8601 ) string.

Answer 4

A regular expression will probably be quite efficient for this:

x <- '2008-08-07T17:07:36Z'
x
## [1] "2008-08-07T17:07:36Z"
sub('.*T(.*)Z', '\\1', x)
## [1] "17:07:36"