An ugly Python script to work out what proportion of weekday instances are on odd "day of month" numbers for a given year

examples.txt

$ python3 odd_days.py 2012
Mon	0.49056603773584906
Tues	0.5192307692307693
Wed	0.4807692307692308
Thurs	0.5192307692307693
Fri	0.5
Sat	0.5
Sun	0.5094339622641509

$ python3 odd_days.py 2013
Mon	0.5
Tues	0.49056603773584906
Wed	0.5192307692307693
Thurs	0.5
Fri	0.5192307692307693
Sat	0.5
Sun	0.5

odd_days.py

#!/usr/bin/env python3
import sys
import datetime
from collections import defaultdict

def generate_year(year):
    delta = datetime.timedelta(days=1)
    date = datetime.date(year, 1, 1)

    while date.year == year:
        yield date
        date = date + delta

day_numbers = defaultdict(list)

for date in generate_year(int(sys.argv[1])):
    day_numbers[date.isoweekday()].append(int(date.strftime('%m')))

day_odd_prob = {}
for day, nums in day_numbers.items():
    odd = 0
    even = 0
    for n in nums:
        if n % 2 == 0:
            even += 1
        else:
            odd += 1
    day_odd_prob[day] = odd / (odd + even)

DAYS = ('Mon', 'Tues', 'Wed', 'Thurs', 'Fri', 'Sat', 'Sun')

for day in sorted(day_odd_prob.keys()):
    print('{}\t{}'.format(DAYS[day-1], day_odd_prob[day]))