How can I determine the weekday of a leap day?-CodePudding

I am trying to figure out what day of the week the leap day falls under (ex. Sunday, Monday, etc) for each year (ex. 1972 Tuesday) in a range of years.

The code below checks that each year is a leap year and appends it to an array:

import array as year

LeapYear = []

startYear = 1970
endYear = 1980

for year in range(startYear, endYear):
    if (0 == year % 4) and (0 != year % 100) or (0 == year % 400):
        LeapYear.append(year)
        print(year)

CodePudding user response：

You are kind of reinventing the wheel. You can use calendar.isleap to determine if a year is a leap year. To get the zero-indexed day of the week (with the week starting on Mondays), you can use calendar.weekday.

import calendar

days = 'Mon Tue Wed Thu Fri Sat Sun'.split()
start = 1994
end = 2025

for year in range(start, end):
    if calendar.isleap(year):
        day_ix = calendar.weekday(year, 2, 29)
        print(year, days[day_ix])

prints:

1996 Thu
2000 Tue
2004 Sun
2008 Fri
2012 Wed
2016 Mon
2020 Sat
2024 Thu

CodePudding user response：

You can use datetime.date to a) check if Feb 29 exists and b) get its weekday name.

from datetime import date

for year in range(1970, 1980):
    try:
        leap_day = date(year, 2, 29)
    except ValueError:
        # Not a leap year
        continue
    weekday = leap_day.strftime('%A')
    print(year, weekday)

Output:

1972 Tuesday
1976 Sunday

Here I'm using EAFP style, but LBYL would also work perfectly fine, i.e.

if calendar.isleap(year):
    leap_day = date(year, 2, 29)
    ...