Calculate the weekday of any given date

weekday.py

# Calculate the weekday of any given date
# Author: djwatt5

# We use the floor method from the math module
# to calculate the year offset.
import math

# Format: day: 1-31, month: 1-12, year: 1-
# String option: if not specified, the result
# will be an integer (Sunday: 0, Monday: 1, ...,
# Saturday: 6). If specified, the result will be
# the weekday as a string. To calculate the weekday,
# we look for century, year, month, and day offsets
# and then add them up to get the weekday integer.
def weekday(day, month, year, string = False):
	
	# Month offset:
	# To calculate a month's offset, you must add
	# the previous month's offset to a number.
	# This number = 3 if the previous month has 31 days,
	# 2 if it has 30 days, and 0 if it's february.
	# Finally, apply modulo 7 to the result.
	# Start from january offset (0).
	month_offset = [0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5]
	
	# Check if it's a leap year:
	# Leap years always are multiples of 400. If they
	# are not multiples of 400, then they must be
	# multiples of 4 but not 100. We add the missing day
	# to the day offset when the leap year needs to be
	# added; that is, when it's a leap year and it's
	# already march.
	leap = 0
	if month > 2:
		if year % 400 == 0:
			leap = 1
		elif year % 100 == 0:
			leap = 0
		elif year % 4 == 0:
			leap = 1
	
	# To calculate year and century offsets, we separate
	# the last two numbers from the year to get the year
	# from 0 to 99, and the century is the rest of the
	# string. Notice that the first two parts of the if
	# statement handle exceptions; years with two or just
	# one digit. This must be done, since python can't
	# assign empty values to numbers.
	year = str(abs(year))
	if len(year) == 2:
		century = 0
		year = int(year)
	elif len(year) == 1:
		century = 0
		year = int(year)
	else:
		century = int(year[:-2])
		year = int(year[-2:])
	
	# Calculating the offsets:
	# The day offset is the day modulo 7. We add the leap
	# day if it's missing. The month offset is the one from
	# the array at the top. The year is itself plus the floor
	# of one fourth of itself. The century is the nearest multiple
	# of 4 minus 1, minus the year, and then the double. The last
	# two are casted to int to see an integer result printed.
	a = (day + leap) % 7
	b = month_offset[month - 1]
	c = int(year + math.floor(year / 4)) % 7
	d = int((39 - century) % 4) * 2
	
	# If the string option is passed, the corresponding weekday
	# will be selected. If not, we pass directly to the result.
	# The result is the sum of the offsets, then passed to
	# modulo seven to get a real weekday.
	if string:
		weekday = {0: "Sunday",
			   1: "Monday",
			   2: "Tuesday",
			   3: "Wednesday",
			   4: "Thursday",
			   5: "Friday",
			   6: "Saturday"}
		return weekday[(a + b + c + d) % 7]
	return (a + b + c + d) % 7

# Test:
print weekday(22, 5, 2013, string = True) # >>> Wednesday