Time difference in years with lubridate?

Question

I would like to use lubridate to calculate age in years given their date of birth and today's date. Right now I have this:

library(lubridate)
today<-mdy(08312015)
dob<-mdy(09071982)
today-dob

which gives me their age in days.

Answer 1

This is the lubridate approach I would take:

interval(dob, today) / years(1)

Yields the answer of 32 years.

Note that the function will complain that it cannot express the remainder of the fraction of the year. This is because year is not a fixed concept, i.e. 366 in leap years and 365 in non-leap years. You can get an answer with more detail in regard to the number of weeks and days:

interval_period = interval(dob, today)
full_year = interval_period %/% years(1)
remaining_weeks = interval_period %% years(1) %/% weeks(1)
remaining_days = interval_period %% years(1) %% weeks(1) %/% days(1)
sprintf('Your age is %d years, %d weeks and %d days', full_year, remaining_weeks, remaining_days)
# [1] "Your age is 32 years, 51 weeks and 1 days"

Note that I use %/% for division and %% as modulo to get the remaining weeks/days after subtracting the full years/weeks.

Answer 2

as.period(today - dob, unit = "years")

This will give a message that it's only an estimate because it doesn't take into account the exact starting date and end date.

Answer 3

as.duration(interval(dob,today)) %/% as.duration(years(1))

should do the job without errors.