Performance deep dive

In this article, we'll find out how to most efficiently calculate if a year is a leap year in Python.

The approaches page lists two idiomatic approaches to this exercise:

1. Using the boolean chain
2. Using the ternary operator

For our performance investigation, we will also include a two further approaches: 3. datetime addition 4. The calendar.isleap() function from the calendar module in the standard library

Benchmarks

To benchmark the approaches, we wrote a small benchmark application using the timeit module. All methods are "fast", but the difference may be easier to see graphically. Note: The y-axis values in the chart have a 1e-7 multiplier. All run times are sub-microsecond.

Timings for approaches by input year

• The if-statements (boolean chain) is the fastest approach when testing a year that is not evenly divisible by 100 and is not a leap year. Since most years fit those conditions, it is overall the most efficient approach.
• The ternary operator is faster in benchmarking when the year is a leap year or is evenly divisible by 100, but those are the least likely conditions.
• Adding to the datetime may not only be a "cheat", but it is slower than the other approaches.
  • Comparing import datatime and from datetime import datetime, timedelta showed little speed difference (data not shown).
• Using the built-in calendar.isleap() function is terse, convenient and very readable, but not quite as fast as writing your own logic. This is likely due to the overhead of both loading the calendar module and then calling calendar.isleap().

Often, it is helpful to the programmer to use imported packages, but a large import to use a simple function may not give the fastest code. Consider the context, and decide which is best for you in each case.