I've been thinking about this and have been dissatisfied with the two answers so far. I like using lubridate
, as @KFB did, but I also want things wrapped up nicely in a function, as in my answer using the eeptools
package. So here's a wrapper function using the lubridate interval method with some nice options:
#' Calculate age
#'
#' By default, calculates the typical "age in years", with a
#' code{floor} applied so that you are, e.g., 5 years old from
#' 5th birthday through the day before your 6th birthday. Set
#' code{floor = FALSE} to return decimal ages, and change code{units}
#' for units other than years.
#' @param dob date-of-birth, the day to start calculating age.
#' @param age.day the date on which age is to be calculated.
#' @param units unit to measure age in. Defaults to code{"years"}. Passed to link{code{duration}}.
#' @param floor boolean for whether or not to floor the result. Defaults to code{TRUE}.
#' @return Age in code{units}. Will be an integer if code{floor = TRUE}.
#' @examples
#' my.dob <- as.Date('1983-10-20')
#' age(my.dob)
#' age(my.dob, units = "minutes")
#' age(my.dob, floor = FALSE)
age <- function(dob, age.day = today(), units = "years", floor = TRUE) {
calc.age = interval(dob, age.day) / duration(num = 1, units = units)
if (floor) return(as.integer(floor(calc.age)))
return(calc.age)
}
Usage examples:
> my.dob <- as.Date('1983-10-20')
> age(my.dob)
[1] 31
> age(my.dob, floor = FALSE)
[1] 31.15616
> age(my.dob, units = "minutes")
[1] 16375680
> age(seq(my.dob, length.out = 6, by = "years"))
[1] 31 30 29 28 27 26
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