Boolean chain

Leap
Leap in Python 
def leap_year(year):
    return year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)

This might be considered the "most idiomatic" or "most Pythonic" solution, as it is exactly the same as the code implemented by the maintainers of the Python language for the calendar.isleap() method.

The first boolean expression uses the modulo operator to check if the year is evenly divided by 4.

  • If the year is not evenly divisible by 4, then the chain will short circuit due to the next operator being a logical AND {and), and will return False.
  • If the year is evenly divisible by 4, then the year is checked to not be evenly divisible by 100.
  • If the year is not evenly divisible by 100, then the expression is True and the interpreter will stop the evaluation to return True, since the next operator is a logical OR (or).
  • If the year is evenly divisible by 100, then the expression is False, and the returned value from the chain will be if the year is evenly divisible by 400.
year year % 4 == 0 year % 100 != 0 year % 400 == 0 is leap year
2020 True True not evaluated True
2019 False not evaluated not evaluated False
2000 True False True True
1900 True False False False

The chain of boolean expressions is efficient, as it proceeds from testing the most to least likely conditions. It is the fastest approach when testing a year that is not evenly divisible by 100 and is not a leap year.

Operator precedence

The implementation contains one set of parentheses, around the or clause:

  • One set is enough, because the % operator is highest priority, then the == and != relational operators.
  • Those parentheses are required, because and is higher priority than or. In Python, a and b or c is interpreted as (a and b) or c, which would give the wrong answer for this exercise.

If in doubt, it is always permissible to add extra parentheses for clarity.

Refactoring

By using the falsiness of 0, the not operator can be used instead of comparing equality to 0. For example:

def leap_year(year):
    return not year % 4 and (year % 100 != 0 or not year % 400)

It can be thought of as the expression not having a remainder.

8th May 2024 · Found it useful?

Other Approaches to Leap in Python

Other ways our community solved this exercise
import datetime


def leap_year(year):
    return (datetime.datetime(year, 2, 28)
            + datetime.timedelta(days=1)).day == 29
datetime addition

Use datetime addition. 

from calendar import isleap

def leap_year(year):
    return isleap(year)
calendar.isleap() function

Use the calendar module.