On Mar 18, 2013, at 8:05 PM, Ben Tupper wrote:
> 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
>>
> julian(strptime(x1,"%m/%d/%Y %H:%M"))
Time differences in days
[1] 13638.38 13878.58 13881.02
attr(,"tzone")
[1] ""
attr(,"origin")
[1] "1970-01-01 GMT"
And a more ham-handed way:
> as.numeric(strptime(x1,"%m/%d/%Y %H:%M"))/(24*3600)
[1] 13638.38 13878.58 13881.02
>
> 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
>
David Winsemius
Alameda, CA, USA
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