year-quarter-day-setters {clock}

Setters: year-quarter-day

Description

These are year-quarter-day methods for the setter generics.

• set_year() sets the fiscal year.

• set_quarter() sets the fiscal quarter of the year. Valid values are in the range of ⁠[1, 4]⁠.

• set_day() sets the day of the fiscal quarter. Valid values are in the range of ⁠[1, 92]⁠.

• There are sub-daily setters for setting more precise components.

Usage

## S3 method for class 'clock_year_quarter_day'
set_year(x, value, ...)

## S3 method for class 'clock_year_quarter_day'
set_quarter(x, value, ...)

## S3 method for class 'clock_year_quarter_day'
set_day(x, value, ...)

## S3 method for class 'clock_year_quarter_day'
set_hour(x, value, ...)

## S3 method for class 'clock_year_quarter_day'
set_minute(x, value, ...)

## S3 method for class 'clock_year_quarter_day'
set_second(x, value, ...)

## S3 method for class 'clock_year_quarter_day'
set_millisecond(x, value, ...)

## S3 method for class 'clock_year_quarter_day'
set_microsecond(x, value, ...)

## S3 method for class 'clock_year_quarter_day'
set_nanosecond(x, value, ...)

Arguments

x

⁠[clock_year_quarter_day]⁠

A year-quarter-day vector.

value

⁠[integer / "last"]⁠

The value to set the component to.

For set_day(), this can also be "last" to adjust to the last day of the current fiscal quarter.

...

These dots are for future extensions and must be empty.

Value

x with the component set.

Examples

library(magrittr)

# Quarter precision vector
x <- year_quarter_day(2019, 1:4)
x

# Promote to day precision by setting the day
x <- set_day(x, 1)
x

# Or set to the last day of the quarter
x <- set_day(x, "last")
x

# What year-month-day is this?
as_year_month_day(x)

# Set to an invalid day of the quarter
# (not all quarters have 92 days)
invalid <- set_day(x, 92)
invalid

# Here are the invalid ones
invalid[invalid_detect(invalid)]

# Resolve the invalid dates by choosing the previous/next valid moment
invalid_resolve(invalid, invalid = "previous")
invalid_resolve(invalid, invalid = "next")

# Or resolve by "overflowing" by the number of days that you have
# gone past the last valid day
invalid_resolve(invalid, invalid = "overflow")

# This is similar to
days <- get_day(invalid) - 1L
invalid %>%
  set_day(1) %>%
  as_naive_time() %>%
  add_days(days) %>%
  as_year_quarter_day()