Parsing ISO8601 date and time format in R [duplicate]

Answer 1

strptime("2013-04-05 07:49:54-07:00", "%Y-%m-%d %H:%M:%S", tz="UTC") gives 2013-04-05 07:49:54 UTC

Try

timep <- strptime(timenoT, "%Y-%m-%d %H:%M:%S", tz="UTC")

Answer 2

You don't the string replacement.

NA just means that the whole did not work, so do it pieces to build your expression:

R> strptime("2013-04-05T07:49:54-07:00", "%Y-%m-%d") 
[1] "2013-04-05"
R> strptime("2013-04-05T07:49:54-07:00", "%Y-%m-%dT%H:%M") 
[1] "2013-04-05 07:49:00"
R> strptime("2013-04-05T07:49:54-07:00", "%Y-%m-%dT%H:%M:%S")
[1] "2013-04-05 07:49:54" 
R>

Also, for reasons I never fully understood -- but which probably reside with C library function underlying it, %z only works on output, not input. So your NA mostly likely comes from your use of %z.

Answer 3

%z is the signed offset in hours, in the format hhmm, not hh:mm. Here's one way to remove the last :.

newstring <- gsub("(.*).(..)$","\\1\\2","2013-04-05T07:49:54-07:00")
(timep <- strptime(newstring, "%Y-%m-%dT%H:%M:%S%z", tz="UTC"))
# [1] "2013-04-05 14:49:54 UTC"

Also note that you don't have to remove the "T".