[R] julian day form POSIXt object

[Gabor Grothendieck]

On Fri, Sep 30, 2011 at 4:47 AM, maxbre <mbressan at arpa.veneto.it> wrote:
> here is an alternative long and winded solution to the problem sticking on
> the julian function of chron package
>
> test$year <- as.integer(as.character(test$date, "%Y"))
> test$month <- as.integer(as.character(test$date, "%m"))
> test$day <- as.integer(as.character(test$date, "%d"))
> test$hour <- as.integer(as.character(test$date, "%H"))
>
> test$jday <-julian(test$month,test$day,test$year,origin=c(month=1, day=0,
> year=test$year))

Assume:

library(chron)
## test data
d <- as.chron(c("2000-01-01", "2000-02-01"))

## Then here are a few more approaches:

## 1
with(month.day.year(d), julian(month, day, year, origin = c(1, 0, year)))

## 2 - yearmon represents a year and month as a year + month/12 where
## month is 0 for jan, 1 for feb, etc.
library(zoo)
as.numeric(d - as.chron(trunc(as.yearmon(d))) + 1)

## 3
as.numeric(format(as.Date(d), "%j"))

Or if d is one or more "Date" class dates then #1 still works.  #2
works if we replace as.chron with as.Date.  #3 can be reduced to
as.numeric(format(d, "%j")) and does not require any external
packages.




or replace as.chron with as.Date if d is one or more "Date" class dates.

>
> thank you all for helping me in sorting out the problem
>
> by the way I like very much this very compact solution
> test$doy <- as.POSIXlt(test$date)$yday+1
> but on the other hand I'm still struggling to fully understand it,
> especially for what is concerning the part after $: so what is $yday for?
>
> if someone would be so kind to explain me this part I would be very grateful
>
> a newbie in R
> lesson learned
>
> max
>
>
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