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rdatetime
Gregor Thomas
Gregor Thomas

Reputation: 145965

How do you convert POSIX date to day of year?

The title has it: how do you convert a POSIX date to day-of-year?

Upvotes: 61

Views: 51978

Answers (5)

andschar
andschar

Reputation: 3983

The data.table package also provides a yday() function.

require(data.table)
today <- Sys.time()
yday(today)

Upvotes: 2

Dave X
Dave X

Reputation: 5137

I realize it isn't quite what the poster was looking for, but I needed to convert POSIX date-times into a fractional day of the year for time series analysis and ended up doing this:

today <- Sys.time()

doy2015f<-difftime(today,as.POSIXct(as.Date("2015-01-01 00:00", tzone="GMT")),units='days')

Upvotes: 4

Tomas
Tomas

Reputation: 59525

This is the way how I do it:

as.POSIXlt(c("15.4", "10.5", "15.5", "10.6"), format = "%d.%m")$yday
# [1] 104 129 134 160

Upvotes: 2

Gregor Thomas
Gregor Thomas

Reputation: 145965

As ?POSIXlt reveals, a $yday suffix to a POSIXlt date (or even a vector of such) will convert to day of year. Beware that POSIX counts Jan 1 as day 0, so you might want to add 1 to the result.

It took me embarrassingly long to find this, so I thought I'd ask and answer my own question.

Alternatively, the excellent lubridate package provides the yday function, which is just a wrapper for the above method. It conveniently defines similar functions for other units (month, year, hour, ...).

today <- Sys.time()
yday(today)

Upvotes: 38

Gavin Simpson
Gavin Simpson

Reputation: 174853

An alternative is to format the "POSIXt" object using strftime():

R> today <- Sys.time()
R> today
[1] "2012-10-19 19:12:04 BST"
R> doy <- strftime(today, format = "%j")
R> doy
[1] "293"
R> as.numeric(doy)
[1] 293

which is preferable to remembering that the day of the years is zero-based in the POSIX standard.

Upvotes: 77

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