Is there an inverse of the yday lubridate function? - r

Question

I have a list of days in the format "day of the year" obtained by applying lubridate::yday() function to a list of dates. For instance, from the following dates (mm-dd-yyyy format):
01-01-2015
01-02-2015
...
by applying yday() you get
1
2
...
Is there a function that can do the reverse given the yday output and the year? Ie, from a yday value and a year, get back to a date in the mm-dd-yyyy format?

Answer 1

To do this with the lubridate package, you use the parse_date_time() function with j as the order = argument.
Example:
my.yday <- "1"
parse_date_time(x = my.yday, orders = "j")
# [1] "2021-01-01 UTC"
The default appears to be the current year. If you want to specify the year just add it in!
my.yday <- "1"
parse_date_time(x = paste(1995, my.yday), orders = "yj")
# [1] "1995-01-01 UTC"
Note: make sure you supply the julian day as character, not numeric. If numeric it will fail when over 3 digits (weird bug that they don't plan to fix)

Answer 2

Any sequence added to a Date() type creates a new Date() sequence with just that offset.
Witness:
R> as.Date("2016-01-01") + 0:9
[1] "2016-01-01" "2016-01-02" "2016-01-03"
[4] "2016-01-04" "2016-01-05" "2016-01-06"
[7] "2016-01-07" "2016-01-08" "2016-01-09"
[10] "2016-01-10"
R> as.Date("2016-01-01") + 100:109
[1] "2016-04-10" "2016-04-11" "2016-04-12"
[4] "2016-04-13" "2016-04-14" "2016-04-15"
[7] "2016-04-16" "2016-04-17" "2016-04-18"
[10] "2016-04-19"
R>
So once again a so-called lubridate question as nothing to do with that package but simply requires to know how the base R types function.

Answer 3

> yday ("1990-03-17") - 1 + as.Date ("1990-01-01")
[1] "1990-03-17"

Answer 4

Just as round out the answer by #DirkEddelbuettel, since I was having the same issue and its been 3 years.
In order to get the desired mm-dd-yyyy for a given dataset where you have only the year and the calendar day as a result of lubridate::yday().
You just need to offset a Date() type object for that given year starting at january 1st, by what your yday() output is subtracted by one.
Then if you want to get the month or day within that month you can use lubridate's month() and day() functions to pull those parts.
(I assume you have the year since that will impact the calendar date b/c of leap years, messing up your month/day assignment. If not, any year will do)
library(dplyr)
library(magrittr)
#Example dataset with years on/around leap year
my_df <- data.frame(
year = c(2010, 2011, 2012, 2013, 2014, 2015),
my_yday = c(150, 150, 150, 150, 150, 150)
)
#skip straight back to the yyyy-mm-dd format
my_df %>% mutate(new_date = as.Date(paste0(year, "-01-01")) + (my_yday - 1))
#Get every component
my_df %>%
mutate(
new_day = lubridate::day(as.Date(str_c(year, "-01-01")) + (my_yday - 1)),
new_month = lubridate::month(as.Date(paste0(year, "-01-01")) + (my_yday - 1)),
new_date = as.POSIXct(str_c(new_month, new_day, year, sep = "/"),
format = "%m/%d/%y"))