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R package constructing time objects from date and hour(integer)

I have data provided in the form of a date telling the day (format "YYYY-MM-DD", e.g. "2015-03-11" and the hours of the day numbered (0-23).

What is the most convenient way to produce time objects of the form

"2015-03-11" and hour = 0 ->  "2015-03-11 00:00"
"2015-03-11" and hour = 1 ->  "2015-03-11 01:00"
"2015-03-11" and hour = 2 ->  "2015-03-11 02:00"

I could use the Date function from Base or something from xts or timeDate. Should be easy but I am sure someone out there knows it quickly.

EDIT: the data is provided in 2 columns, one for the date and one numerical.

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Richi W Avatar asked Mar 13 '15 10:03

Richi W


2 Answers

You don't need an external package to do that.
If your data is in this format:

df=data.frame(date=c("2015-03-11","2015-03-11","2015-03-11"),hour=0:2)

just apply the following function:

format(as.POSIXct(df$date)+df$hour*60*60, format = "%Y-%m-%d %H:%M")
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RockScience Avatar answered Sep 22 '22 02:09

RockScience


Suppose we have this input:

date <- c("2015-03-11", "2015-03-12")
hour <- 2:3

then try one of these:

1) chron

library(chron)
as.chron(date) + hour/24

giving:

[1] (03/11/15 02:00:00) (03/12/15 03:00:00)

2) POSIXct. This one only uses the base of R, no packages:

as.POSIXct(date) + 3600 * hour    

giving, on my system:

[1] "2015-03-11 02:00:00 EDT" "2015-03-12 03:00:00 EDT"

If you wanted the result in the UTC time zone use:

as.POSIXct(date, tz = "UTC") + 3600 * hour  

3) lubridate

library(lubridate)
ymd(date) + hours(hour)

giving:

[1] "2015-03-11 02:00:00 UTC" "2015-03-12 03:00:00 UTC"

If you want it in the current time zone then:

ymd(date, tz = "") + hours(hour)

Note that the chron solution gives a date/time class that does not use time zones eliminating the many problems that time zones can cause. The POSIXct and lubridate solutions give the date/time in a specific time zone as shown.

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G. Grothendieck Avatar answered Sep 25 '22 02:09

G. Grothendieck