HI,
The question is not clear. If it is to get the hours,
strptime(dates, format=%d/%m/%Y %H:%M)$hour
# [1] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 0
#or
as.numeric(format(as.POSIXct(dates, format=%d/%m/%Y %H:%M),%H))
# [1] 0 1 2 3 4 5 6 7 8 9 10 11
Why am I know getting hours after I convert the date?
dates - c('31/12/2013 0:00', '31/12/2013 1:00', '31/12/2013 2:00',
'31/12/2013 3:00', '31/12/2013 4:00', '31/12/2013 5:00', '31/12/2013
6:00', '31/12/2013 7:00', '31/12/2013 8:00', '31/12/2013 9:00',
'31/12/2013 10:00', '31/12/2013
as.Date produces Dates only, with no time information, even if you try
to supply it with hours + minutes.
For dates+times, use as.POSIXct() or as.POSIXlt() in place of
as.Date(). POSIXct produces a numeric value for the number of seconds
since your specified origin time (usually 1970-01-01
Because a Date object represents calendar dates, and calendar dates don't
have hours.
Use as.POSIXct() instead of as.Date()
(and spend a little more time with the documentation for as.Date)
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
Hi,
You could use:
strptime(dates, format=%d/%m/%Y %H:%M)[1:2]
#[1] 2013-12-31 00:00:00 2013-12-31 01:00:00
as.POSIXlt(dates, format=%d/%m/%Y %H:%M)[1:2]
#[1] 2013-12-31 00:00:00 2013-12-31 01:00:00
A.K.
Thanks. as.POSIXct works for the most part. The only problem is part of
data I'm working
This function returns date/times without timezone
strptime(dates,format=%d/%m/%Y %H:%M)
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Thanks. as.POSIXct works for the most part. The only problem is part of data
I'm working has it's own time zone. Is there a way to not have a time zone
displayed? My times do not change with Daylight saving.
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You must deal with identifying the time zone. I have found that setting TZ
environment variable appropriately for the data before converting character
values to POSIXct gives me the best results. This is actually easier for
standard-time-only data than for data with daylight savings time
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