How do I identify leap years in Stata?
Leap year indicators
Nicholas J. Cox, Durham University, UK
How do you determine if a year is a leap year using Stata syntax? For
example, you know that 2004 and 2008 are leap years, but how would you
handle leap years generally? Even if you are not especially interested in
leap years, the answer provides good examples of some key Stata functions,
so do read on.
The rules in the Gregorian calendar for a year to be a leap year are
YES if 1. year divisible by 4
(but NO if 2. year divisible by 100
(but YES if 3. year divisible by 400))
and NO otherwise.
Note the nesting of rules. If a leap year is a first-order correction, the
third rule is an example of a third-order correction. Scientists and
engineers often use such ideas, but they seem in shorter supply in the
You will be familiar with rule 1, but rules 2 and 3 are occasionally
forgotten. For example, Excel has 1900 as a leap year; it is documented that
this was to provide compatibility with Lotus 1-2-3.
In Stata, suppose that year is a variable. An indicator containing 1
for leap year and 0 otherwise is then given by
(mod(year,4) == 0 & mod(year,100) != 0) | mod(year,400) == 0
as mod(,) provides the remainder left over from division.
Alternatively, we could use cond(,,):
cond(mod(year,400) == 0, 1,
cond(mod(year,100) == 0, 0,
cond(mod(year,4) == 0, 1,
The layout of such code may be important to you, if not to Stata. In such
expressions, to make sure you have balanced parentheses, exploit the
pertinent function in a decent text editor. In the Stata Do-file Editor, it
is "Balance", Ctrl-B; in Vim, it is the % key; etc.
Given daily dates, a feature of a leap year is clearly that there is a
February 29, so
mdy(2,29,2004) < .
mdy(2,29,2004) != .
is true. That is, mdy(2,29,whenever) is missing if
whenever is not a leap year but is nonmissing otherwise; in the
latter case, it is less than missing. You could also do that as
Similarly, there are 366 days in a leap year, so
doy(mdy(12,31,2004)) == 366
is true. We could vectorize both of those calculations to say
mdy(2,29,year) < .
doy(mdy(12,31,year)) == 366
Reingold and Dershowitz (2001) provide definitive explanations of
- Reingold, E. M., and N. Dershowitz. 2001.
- Calendrical Calculations: The Millennium Edition.
Cambridge: Cambridge University Press.
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