`period()`

creates a period object with the specified values.

period(num = NULL, units = "second", ...) is.period(x) seconds(x = 1) minutes(x = 1) hours(x = 1) days(x = 1) weeks(x = 1) years(x = 1) milliseconds(x = 1) microseconds(x = 1) nanoseconds(x = 1) picoseconds(x = 1) # S3 method for numeric months(x, abbreviate)

num | a numeric vector that lists the number of time units to be
included in the period. From v1.6.0 |
---|---|

units | a character vector that lists the type of units to be used. The
units in units are matched to the values in num according to their
order. When |

... | a list of time units to be included in the period and their
amounts. Seconds, minutes, hours, days, weeks, months, and years are
supported. Normally only one of |

x | Any R object for |

abbreviate | Ignored. For consistency with S3 generic in base namespace. |

a period object

Within a Period object, time units do not have a fixed length (except for seconds) until they are added to a date-time. The length of each time unit will depend on the date-time to which it is added. For example, a year that begins on 2009-01-01 will be 365 days long. A year that begins on 2012-01-01 will be 366 days long. When math is performed with a period object, each unit is applied separately. How the length of a period is distributed among its units is non-trivial. For example, when leap seconds occur 1 minute is longer than 60 seconds.

Periods track the change in the "clock time" between two date-times. They are measured in common time related units: years, months, days, hours, minutes, and seconds. Each unit except for seconds must be expressed in integer values.

Period objects can be easily created with the helper functions
`years()`

, `months()`

, `weeks()`

,
`days()`

, `hours()`

, `minutes()`

,
and `seconds()`

. These objects can be added to and subtracted
to date-times to create a user interface similar to object oriented programming.

Note: Arithmetic with periods can results in undefined behavior when
non-existent dates are involved (such as February 29th). Please see
Period for more details and `%m+%`

and
`add_with_rollback()`

for alternative operations. Note: Arithmetic with
periods can results in undefined behavior when non-existent dates are
involved (such as February 29th in non-leap years). Please see Period
for more details and `%m+%`

and `add_with_rollback()`

for
alternative operations.

Period, `period()`

, `%m+%`

,
`add_with_rollback()`

period(c(90, 5), c("second", "minute"))#> [1] "5M 90S"# "5M 90S" period(-1, "days")#> [1] "-1d 0H 0M 0S"period(c(3, 1, 2, 13, 1), c("second", "minute", "hour", "day", "week"))#> [1] "20d 2H 1M 3S"period(c(1, -60), c("hour", "minute"))#> [1] "1H -60M 0S"period(0, "second")#> [1] "0S"period (second = 90, minute = 5)#> [1] "5M 90S"period(day = -1)#> [1] "-1d 0H 0M 0S"period(second = 3, minute = 1, hour = 2, day = 13, week = 1)#> [1] "20d 2H 1M 3S"period(hour = 1, minute = -60)#> [1] "1H -60M 0S"period(second = 0)#> [1] "0S"period(c(1, -60), c("hour", "minute"), hour = c(1, 2), minute = c(3, 4))#> [1] "1H -60M 0S" "1H 3M 0S" "2H 4M 0S"period("2M 1sec")#> [1] "2M 1S"period("2hours 2minutes 1second")#> [1] "2H 2M 1S"period("2d 2H 2M 2S")#> [1] "2d 2H 2M 2S"period("2days 2hours 2mins 2secs")#> [1] "2d 2H 2M 2S"#> [1] "172800s (~2 days)"#> [1] TRUE#> [1] "2009-08-04 06:30:00 UTC"x + days(100) - hours(8)#> [1] "2009-11-10 16:00:00 UTC"class(as.Date("2009-08-09") + days(1)) # retains Date class#> [1] "Date"as.Date("2009-08-09") + hours(12)#> [1] "2009-08-09 12:00:00 UTC"class(as.Date("2009-08-09") + hours(12))#> [1] "POSIXlt" "POSIXt"# converts to POSIXt class to accomodate time units years(1) - months(7)#>#> #>#> [1] "1y -7m 0d 0H 0M 0S"c(1:3) * hours(1)#> [1] "1H 0M 0S" "2H 0M 0S" "3H 0M 0S"hours(1:3)#> [1] "1H 0M 0S" "2H 0M 0S" "3H 0M 0S"#> [1] "2009-01-01" "2009-02-01" "2009-03-01" "2009-04-01" "2009-05-01" #> [6] "2009-06-01" "2009-07-01" "2009-08-01" "2009-09-01" "2009-10-01" #> [11] "2009-11-01" "2009-12-01"# compare DST handling to durations boundary <- ymd_hms("2009-03-08 01:59:59", tz="America/Chicago") boundary + days(1) # period#> [1] "2009-03-09 01:59:59 CDT"#> [1] "2009-03-09 02:59:59 CDT"is.period(as.Date("2009-08-03")) # FALSE#> [1] FALSEis.period(period(months= 1, days = 15)) # TRUE#> [1] TRUE