Hi Janesh,
On Mar 18, 2013, at 10:49 PM, Janesh Devkota wrote:
> Dear R Users,
>
> I have data for more than 3 years. For each year I want to find the day
> corresponding to Jaunary 1 of that year. For example:
>
>> x <- c('5/5/2007','12/31/2007','1/2/2008')
>
>> #Convert to day of year (julian date) -
>
>> strptime(x,"%m/%d/%Y")$yday+1
>
> [1] 125 365 2
>
> I want to know how to do the same thing but with time added. But I still get
> the day not time. Can anyone suggest what is the better way to find the
> julian date with date and time ?
>
>> x1 <- c('5/5/2007 02:00','12/31/2007 05:58','1/2/2008 16:25')
>
>> #Convert to day of year (julian date) -
>
>> strptime(x1,"%m/%d/%Y %H:%M")$yday+1
>
> [1] 125 365 2
>
Would this do it for you?
x1 <- c('5/5/2007 02:00','12/31/2007 05:58','1/2/2008 16:25')
x1 <- as.POSIXct(x1, format = "%m/%d/%Y %H:%M")
day <- as.numeric(format(x1, "%j"))
hour <- as.numeric(format(x1, "%H"))/24
minute <- as.numeric(format(x1, "%M"))/(60*24)
second <- as.numeric(format(x1, "%S"))/(60*60*24)
day + hour + minute + second
[1] 125.083333 365.248611 2.684028
Cheers,
Ben
> Thank you so much.
>
> Janesh
>
>
> [[alternative HTML version deleted]]
>
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Ben Tupper
Bigelow Laboratory for Ocean Sciences
60 Bigelow Drive, P.O. Box 380
East Boothbay, Maine 04544
http://www.bigelow.org
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