Convert day of year to date

Question

How do I convert day-of-year to date? For example, what's the date for 104th day of 2004?

This is the opposite of How do you convert POSIX date to day of year in R?

Inelegancy

Answer 1

I found that @TMS answer ignores the year specified in the origin, replacing it with the actual year you are executing the command (as @Shekeine found). Doing this seems to work fine though:

as.Date(104, origin = "2014-01-01")

CAUTION: day-of-year is zero based in this case!

Satinet

Thank you for this. I am not sure why my code no longer works (I think it was working back then). Added an important warning to your answer as well. – Inelegancy

As noted above, that code gives a zero based day of year (ie, Jan 1st is DOY=0). If you want the more typical 1-based day of year (ie, Jan 1st is DOY=1) you'd use origin="2013-12-31" – Poison

Year is hardcoded, which makes it pretty much useless for dates that are not 2014. If year is a variable in your code, and you have multiple years, see @BonnieM answer below. – Ber

Answer 2

This is the way I do it:

as.Date(104, format = "%j", origin = "1.1.2014")
# "2014-04-15"

PS: for those who wonder if answering my own question is encouraged, please look here: https://meta.stackexchange.com/questions/17463/can-i-answer-my-own-questions-even-if-i-knew-the-answer-before-asking

Inelegancy

Hi, your code above returns [1] "2015-04-15" for me. Running as.Date(1, format = "%j", origin=paste0("1.1.", "2015")) returns "2015-01-02". What am I missing here? – Escritoire

I am not sure why my code no longer works (I think it was working back then). I accepted the other answer. – Inelegancy

@Escritoire I didn't understand what was wrong with your output. If you're concerned that it returns 2nd of January - I think the counting starts at 0 in this case. – Explanatory

Answer 3

For anyone working with many years of DOY data, here's an easy way to extract the date, month, and day from DOY using dplyr.

library(dplyr)
#
# create example dataframe 
x <- data.frame(doy=c(rep(1:5, 3)), 
                year=c(rep(2010, 5), rep(2011, 5), rep(2012, 5)))
#
# change doy to dates using mutate()
x %>% mutate(date_= as.Date(doy-1, origin=paste0(year, "-01-01")), 
              month= strftime(date_, "%m"), 
              day=strftime(date_,"%d"))  
# subtract 1 from doy because R uses a 0 base index 
# otherwise DOY=1 will be Jan. 2.
# output:
    doy year      date_ month day
1    1 2010 2010-01-01    01  01
2    2 2010 2010-01-02    01  02
3    3 2010 2010-01-03    01  03
4    4 2010 2010-01-04    01  04
5    5 2010 2010-01-05    01  05
6    1 2011 2011-01-01    01  01
7    2 2011 2011-01-02    01  02
8    3 2011 2011-01-03    01  03
9    4 2011 2011-01-04    01  04
10   5 2011 2011-01-05    01  05
11   1 2012 2012-01-01    01  01
12   2 2012 2012-01-02    01  02
13   3 2012 2012-01-03    01  03
14   4 2012 2012-01-04    01  04
15   5 2012 2012-01-05    01  05

Afterthought

By far the best answer because it doesn't depend on the year. The accepted answer has the year hardcoded in. My df contains many years and this answer works. – Ber

Answer 4

as.Date(2, origin = '2014-01-01')

As for the issue with above code giving the output as 2014-01-03.

You can just put the origin as the last day of the previous year.

as.Date(2, origin = '2013-12-31')

It works that way.

Chenab