Extract time from timestamp?

问题:

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 different times of day and day and date is irrelevant.

However, this is how the timestamps are structured in the dataset: 2008-08-07T17:07:36Z

And I'm unsure how to only get that time from this timestamp.

Thank you for any help you can provide and please just let me know if I can provide more information!

回答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"

回答2:

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.

回答3:

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"

回答4:

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"