change a column from birth date to age in r

Question

I am using data.table for the first time.

I have a column of about 400,000 ages in my table. I need to convert them from birth dates to ages.

What is the best way to do this?

Answer 1

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