How to account for leap years?

Question

I have some doubts about the leap years, how can I be sure that by using a formula like this

add.years= function(x,y){    
if(!isTRUE(all.equal(y,round(y))))         


        

Answer 1

A year is a leap year if:

• Is divisible by 4.
• Not if it is divisible by 100.
• But is if it is divisible by 400.

That is why 2000 was a leap year (although it's divisible by 100, it's also divisible by 400).

But generally, if you have a library that can take of date/time calculations then use it. It's very complicated to do these calculations and easy to do wrong, especially with ancient dates (calendar reforms) and timezones involved.

Answer 2

You can check if a year is a leap year with leap_year from lubridate.

years <- 1895:2005
years[leap_year(years)]

This package will also handle not generating impossible 29ths of February.

ymd("2000-2-29") + years(1)    # NA
ymd("2000-2-29") %m+% years(1) # "2001-02-28"

The %m+% "add months" operator, as mentioned by @VitoshKa, rolls the date back to the end of the previous month if the actual day doesn't exist.

Answer 3

Following the suggestion of DarkDust and Dirk Eddelbuettel, you can easily roll your own leap_year function:

leap_year <- function(year) {
  return(ifelse((year %%4 == 0 & year %%100 != 0) | year %%400 == 0, TRUE, FALSE))
}

and apply it to vector data:

years = 2000:2050
years[leap_year(years)]

[1] 2000 2004 2008 2012 2016 2020 2024 2028 2032 2036 2040 2044 2048

Answer 4

Your suspicions are indeed correct:

x <- as.POSIXlt("2000-02-29")
y <- x
y$year <- y$year+100
y
#[1] "2100-03-01"

The strange thing is that other parts of y remain unchanged so you can't use these for comparison:

y$mday
#[1] 29
y$mon
#[1] 1

But you can use strftime:

strftime(x,"%d")
#[1] "29"
strftime(y,"%d")
#[1] "01"

So how about:

add.years <- function(x,y){
   if(!isTRUE(all.equal(y,round(y)))) stop("Argument \"y\" must be an integer.\n")
   x.out <- as.POSIXlt(x)
   x.out$year <- x.out$year+y
   ifelse(strftime(x,"%d")==strftime(x.out,"%d"),as.Date(x.out),NA)
   } 

You can then subset your data using [ and is.na to get rid of the otherwise duplicate 1st March dates. Though as these dates seem to be consecutive, you might want to consider a solution that uses seq.Date and avoid dropping data.