These are year-day methods for the arithmetic generics.
Notably, you cannot add days to a year-day. For day-based arithmetic,
first convert to a time point with as_naive_time() or as_sys_time().
Usage
# S3 method for clock_year_day
add_years(x, n, ...)Arguments
- x
[clock_year_day]A year-day vector.
- n
[integer / clock_duration]An integer vector to be converted to a duration, or a duration corresponding to the arithmetic function being used. This corresponds to the number of duration units to add.
nmay be negative to subtract units of duration.- ...
These dots are for future extensions and must be empty.
Details
x and n are recycled against each other using
tidyverse recycling rules.
Examples
x <- year_day(2019, 10)
add_years(x, 1:5)
#> <year_day<day>[5]>
#> [1] "2020-010" "2021-010" "2022-010" "2023-010" "2024-010"
# A valid day in a leap year
y <- year_day(2020, 366)
y
#> <year_day<day>[1]>
#> [1] "2020-366"
# Adding 1 year to `y` generates an invalid date
y_plus <- add_years(y, 1)
y_plus
#> <year_day<day>[1]>
#> [1] "2021-366"
# Invalid dates are fine, as long as they are eventually resolved
# by either manually resolving, or by calling `invalid_resolve()`
# Resolve by returning the previous / next valid moment in time
invalid_resolve(y_plus, invalid = "previous")
#> <year_day<day>[1]>
#> [1] "2021-365"
invalid_resolve(y_plus, invalid = "next")
#> <year_day<day>[1]>
#> [1] "2022-001"
# Manually resolve by setting to the last day of the year
invalid <- invalid_detect(y_plus)
y_plus[invalid] <- set_day(y_plus[invalid], "last")
y_plus
#> <year_day<day>[1]>
#> [1] "2021-365"