On 27/03/2012 19:30, Justin Haynes wrote:
There may very well be a better solution, but this works.
format(strptime(dayofyear, format="%j"), format="%m-%d")
The answer depends on the year (think leap years), so I think you need
strptime(paste("2008", dayofyear), format="%Y %j")
Probably a better idea is
as.Date(dayofyear - 1, origin = "2008-01-01")
(as Jan 1 is day 1).
On Tue, Mar 27, 2012 at 11:12 AM, Sam Albers<tonightstheni...@gmail.com>wrote:
Hello,
I am having trouble figuring out how to convert a Day of Year integer
back into a Date format. For example I have the following:
date<-
c('2008-01-01','2008-01-02','2008-01-03','2008-01-04','2008-01-05','2008-01-06','2008-01-07',
'2008-01-08','2008-01-09','2008-01-10','2008-01-11','2008-01-12','2008-01-13','2008-01-14','2008-01-15',
'2008-01-16','2008-01-17','2008-01-18','2008-01-19','2008-01-20','2008-01-21','2008-01-22','2008-01-23')
## this is then converted into a number corresponding to the day of
the year like so:
dayofyear<- strptime(date, format="%Y-%m-%d")$yday + 1
## Now my question is how do I get back to a date format (obviously
omitting the year).
## The end result is that I'd like to be able to have axis labels as
something like "Month-Day" or just "Month"
## instead of just an integers which isn't always intuitive for people
but I can't seem to figure out how to tell R
## to recognize an integer as a date.
Any suggestions?
Many thanks in advance!
Sam
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--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
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