Parsing ISO8601 date and time format in R

Question

This should be quick - we are parsing the following format in R:

2013-04-05T07:49:54-07:00

My current approach is

require(stringr) 
timenoT <- str_replace_all("2013-04-05T07:49:54-07:00", "T", " ") 
timep <- strptime(timenoT, "%Y-%m-%d %H:%M:%S%z", tz="UTC")

but it gives NA.

`%z` is the signed offset in hours, in the format `hhmm`, not `hh:mm`. — Joshua Ulrich, Apr 05 '13 at 16:08

score 17 · Accepted Answer · answered Apr 05 '13 at 16:14

17

%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".

answered Apr 05 '13 at 16:14

Joshua Ulrich

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That seems like an overly general pattern. I would think something like this would be less prone to over-correct: `gsub("([+-]\\d\\d)(:)", "\\1", argvec)` – IRTFM Apr 05 '13 at 16:45
I need to do exactly the opposite! How can I add `:` in the last part? For me, it always shows `+0200`, but I need `+02:00`. – PM0087 Apr 23 '20 at 10:02
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@PeyM87: something like this should be close: `gsub("(.*)([+-][[:digit:]]{2})([[:digit:]]{2})$", "\\1\\2:\\3", x)` – Joshua Ulrich Apr 23 '20 at 10:39

score 1 · Answer 2 · answered Apr 05 '13 at 16:12

1

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.

answered Apr 05 '13 at 16:12

Dirk Eddelbuettel

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`%z` works on input, but the format is different. `%Z` only works on output. – Joshua Ulrich Apr 05 '13 at 16:16
Ah, yes, thanks. My solution still gets by without a regexp, or Hadley package. So by our definition, I win. – Dirk Eddelbuettel Apr 05 '13 at 16:58
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It does all those things and, as an added bonus, provides the wrong answer. :) – Joshua Ulrich Apr 05 '13 at 17:44

score 0 · Answer 3 · answered Apr 05 '13 at 16:12

0

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")

answered Apr 05 '13 at 16:12

rongenre

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That's incorrect, because the time isn't 7:49:54 UTC, it's 7 hours behind UTC. So if you express it _in_ UTC, it will be 7 hours ahead of 7:49.54. – Joshua Ulrich Apr 05 '13 at 16:23

Parsing ISO8601 date and time format in R

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