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Determine, count, and list leap years in Python

Modified: | Tags: Python, Date and time

In Python, the calendar module provides functions to determine if a year is a leap year and to count leap years in a specified period.

Contents

See the following article on how to create a text or HTML calendar with the calendar module.

The calendar module is included in the standard library, so no additional installation is required.

The algorithm for leap years

The algorithm for determining a leap year in the Gregorian calendar is represented by the following pseudocode:

if (year is not divisible by 4) then (it is a common year)
else if (year is not divisible by 100) then (it is a leap year)
else if (year is not divisible by 400) then (it is a common year)
else (it is a leap year)

Determine if a year is a leap year: calendar.isleap()

Use calendar.isleap() to determine if a year is a leap year.

import calendar

print(calendar.isleap(2019))
# False

print(calendar.isleap(2020))
# True

print(calendar.isleap(1900))
# False

print(calendar.isleap(2000))
# True
source: calendar_leap.py

The source code for calendar.isleap() is as follows, using the modulo operator % and logical operators and and or.

def isleap(year):
    """Return True for leap years, False for non-leap years."""
    return year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)
source: calendar.py

Count leap years in a specified period: calendar.leapdays()

Use calendar.leapdays() to count leap years in a specified period.

This function takes two arguments, y1 and y2, and counts leap years from y1 (inclusive) to y2 (exclusive).

import calendar

print(calendar.leapdays(2019, 2030))
# 3

print(calendar.leapdays(2019, 2020))
# 0
source: calendar_leap.py

List leap years in a specified period

To list leap years in a specified period, use calendar.isleap() as a condition in list comprehension.

import calendar

print([y for y in range(2019, 2030) if calendar.isleap(y)])
# [2020, 2024, 2028]

print([y for y in range(2000, 2020) if calendar.isleap(y)])
# [2000, 2004, 2008, 2012, 2016]
source: calendar_leap.py

Note that in range(start, end), end is not included.

Determine if a datetime or date object is a leap year

Python's standard library includes the datetime module for handling dates and times.

The datetime module provides datetime.datetime for both date and time, and datetime.date for dates only.

Both datetime.datetime and datetime.date have the year attribute. To determine if the year of a datetime.datetime or datetime.date object is a leap year, pass its year attribute to calendar.isleap().

import calendar
import datetime

dt = datetime.datetime(2019, 1, 1, 10, 10, 10)
print(dt)
# 2019-01-01 10:10:10

print(calendar.isleap(dt.year))
# False

d = datetime.date(2020, 1, 1)
print(d)
# 2020-01-01

print(calendar.isleap(d.year))
# True
source: calendar_datetime_leap.py

For example, you can define the following function. The argument can be either datetime.datetime or datetime.date.

def isleap_datetime(dt):
    return calendar.isleap(dt.year)

print(dt)
# 2019-01-01 10:10:10

print(isleap_datetime(dt))
# False

print(d)
# 2020-01-01

print(isleap_datetime(d))
# True
source: calendar_datetime_leap.py

If you prefer not to use the calendar module, here is an alternative way.

def isleap_datetime2(dt):
    return dt.year % 4 == 0 and (dt.year % 100 != 0 or dt.year % 400 == 0)

print(dt)
# 2019-01-01 10:10:10

print(isleap_datetime2(dt))
# False

print(d)
# 2020-01-01

print(isleap_datetime2(d))
# True
source: calendar_datetime_leap.py

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