year-day-arithmetic | R Documentation |
These are year-day methods for the arithmetic generics.
add_years()
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()
.
## S3 method for class 'clock_year_day'
add_years(x, n, ...)
x |
A year-day vector. |
n |
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. |
... |
These dots are for future extensions and must be empty. |
x
and n
are recycled against each other using
tidyverse recycling rules.
x
after performing the arithmetic.
x <- year_day(2019, 10)
add_years(x, 1:5)
# A valid day in a leap year
y <- year_day(2020, 366)
y
# Adding 1 year to `y` generates an invalid date
y_plus <- add_years(y, 1)
y_plus
# 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")
invalid_resolve(y_plus, invalid = "next")
# 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
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