Convert fraction of day to POSIX times in R [duplicate]

Question

I have a dataset that encodes a date-time into two separate variables. Normally, I'd just paste them together inside of an as.POSIXct and carry on. However, the date is provided as a string, and the time of day as a fraction of 24 hours - e.g., 12pm is 0.5, 9:30am is 0.1458333, etc.

It doesn't seem all that tricky to convert the fractional days into clock hours, but I'd prefer to use a pre-existing function if possible. Does something like that exist in base R? A package?

If it's any use, this is an Excel (xlsx) time field imported into R through RODBC.

EDIT Oddly enough, upon revisiting this problem, the times are now read in as POSIXct. Not sure what to make of that.

Answer 1

The R News 4/1 Help Desk article has a section on reading Excel dates in R.

Answer 2

POSIXct values are simply the number of seconds since midnight GMT 1970-01-01. (So you need to pay attention to your offset from UTC.) You can use the date part and add the number of days times 24*3600 (as.Date(dtval) to your time value * 24*3600. Gabor pointed to the article in R News (which he wrote, thank you, Gabor.)

You didn't give an example of the string. If you are getting your date as a string, then as.Date(strDate) will convert a variable "strDate" to Date class when it is in either "YYYY-MM-DD" or "YYYY/MM/DD" format. Otherwise the formatting codes are on the ?strptime page.

Once you have a POSIXct-classed variable you can just add the number of seconds. This example add 30 minutes to midnight today Feb 1, 2011 (in my time zone which is UTC-5):

> as.POSIXct(as.Date("2011-02-01")) +30*60
[1] "2011-01-31 19:30:00 EST"

And this is your time value added to midnight my time:

> as.POSIXct(as.Date("2011-02-01 00:00", tzone="UTC"))+3600*5 + 3600*24*timeval
[1] "2011-02-01 03:29:59 EST"