Dates with lubridate

Reference material — not covered in the taught sessions

Author

Gabriel Mateus Bernardo Harrington

This page is reference material, not a taught session. It is here because dates turn up constantly in real clinical and biological data — sample collection dates, visit dates, dates of birth — and doing arithmetic on them by hand is a reliable way to get a wrong answer.

Setup

library(lubridate)
library(dplyr)
library(ggplot2)

How dates work in R

  • Dates and times are not just strings. They come in many formats — YYYY-MM-DD, MM/DD/YYYY, DD-MM-YYYY.
  • Handling them means dealing with varied formats, time zones and leap years, as well as calculations between dates.
  • R has a dedicated date type, and packages to parse and manage it.
class("2024-10-20")           # a plain string
[1] "character"
class(as.Date("2024-10-20"))  # a date object
[1] "Date"
typeof(as.Date("2024-10-20")) # but a double underneath
[1] "double"

That last line is the thing to remember: a date is stored as a number (days since 1970-01-01). That is what makes arithmetic on dates work at all — and why a date that has been silently read in as a string will fail in confusing ways.

Introduction to lubridate

lubridate makes working with dates and times easier. With it you can:

  • parse dates from strings in common formats
  • do arithmetic with dates easily
  • account for time zones, leap years and so on
  • handle times as well, though that isn’t covered here

It is genuinely not possible to be accurate with dates without a dedicated package. Do not try to do it with substr() and arithmetic.

Parsing and extracting

  • ymd() converts strings in year-month-day order to dates. Related functions parse other orders: mdy(), dmy(), ymd_hms() and so on. You choose the function that matches the order of the parts, not the separators — ymd() copes with 2024-10-20, 2024/10/20 and 20241020 alike.
  • today() returns the current date.
  • year(), month() and day() extract components from a date.
d <- ymd("2024-10-20")
c(year = year(d), month = month(d), day = day(d))
 year month   day 
 2024    10    20

Intervals and durations

interval() creates an interval between two dates, and time_length() measures it in whatever unit you ask for.

start_date <- ymd("2015-05-15")
end_date <- ymd("2024-10-20")
date_interval <- interval(start_date, end_date)

time_length(date_interval, "years")
[1] 9.43287671233

Use time_length(interval(...), "years") rather than dividing a difference by 365. The second approach is wrong by a day roughly every four years, which is exactly the kind of error that survives review because it is small.

Practice

  • John Doe was born on 4th September 1983. Create a date object for his birth date.
  • How old was he on 6th June 2020? What about today?

Using the lakers dataset, which ships with lubridate

  • Choose the correct function to replace some_function below:
lakers |>
  as_tibble() |>
  mutate(date = some_function(sprintf("%08d", date)))

Using the economics dataset from ggplot2

  • Take the date column and create two new columns holding the year and month of each date.
  • Create a column time_since_nyse holding the number of years between the founding of the New York Stock Exchange on 17th May 1792 and each date.

Further reading