Using dateutil.relativedelta
:
from dateutil.relativedelta import relativedelta, FR # $ pip install python-dateutil
def third_friday_dateutil(now):
"""the 3rd Friday of the month, not the 3rd Friday after today."""
now = now.replace(day=1) # 1st day of the month
now += relativedelta(weeks=2, weekday=FR)
return now
Or using dateutil.rrule
:
from datetime import date, timedelta
from dateutil.rrule import rrule, MONTHLY, FR
def third_friday_rrule(now):
return rrule(MONTHLY, count=1, byweekday=FR, bysetpos=3, dtstart=now.replace(day=1))[0]
def get_third_fris_rrule(how_many):
return list(rrule(MONTHLY, count=how_many, byweekday=FR, bysetpos=3, dtstart=date.today()+timedelta(1)))
Here's a brute force solution (15x times faster):
#!/usr/bin/env python
import calendar
from datetime import date, timedelta
from itertools import islice
DAY = timedelta(1)
WEEK = 7*DAY
def fridays(now):
while True:
if now.weekday() == calendar.FRIDAY:
while True:
yield now
now += WEEK
now += DAY
def next_month(now):
"""Return the first date that is in the next month."""
return (now.replace(day=15) + 20*DAY).replace(day=1)
def third_friday_brute_force(now):
"""the 3rd Friday of the month, not the 3rd Friday after today."""
return next(islice(fridays(now.replace(day=1)), 2, 3))
def get_third_fris(how_many):
result = []
now = date.today()
while len(result) < how_many:
fr = third_friday_brute_force(now)
if fr > now: # use only the 3rd Friday after today
result.append(fr)
now = next_month(now)
return result
print(get_third_fris(6))
Output
[datetime.date(2015, 3, 20),
datetime.date(2015, 4, 17),
datetime.date(2015, 5, 15),
datetime.date(2015, 6, 19),
datetime.date(2015, 7, 17),
datetime.date(2015, 8, 21)]
See Converting datetime.date to UTC timestamp in Python
Here's comparison with other solutions and tests (for all possible 400 years patterns).