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.