weekdays in range
(Sorry if this shows up twice.) Can someone think of an easy way to calculate the number of weekdays between two calendar dates (in Python)? Thankee. -- http://mail.python.org/mailman/listinfo/python-list
Re: weekdays in range
I'm using weekdays as a proxy for days when the US stock market is open. (I'll miss holidays.) The application is pricing CALL and PUT options. Speed is not a problem. The number of days will typically be between 1 and 254. -- http://mail.python.org/mailman/listinfo/python-list
Re: weekdays in range
Jive Dadson notonthe...@noisp.com writes: Can someone think of an easy way to calculate the number of weekdays between two calendar dates (in Python)? That depends on what you mean by “weekdays”. import datetime begin_date = datetime.date(2009, 10, 9) end_date = datetime.date(2009, 10, 22) import calendar print calendar.month(2009, 10) October 2009 Mo Tu We Th Fr Sa Su 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 end_date - begin_date datetime.timedelta(13) If you're expecting to exclude Saturday and Sunday (i.e. if you expect the above result to be 9 days instead of 13), you can use other functions of the ‘calendar’ module; try starting with: friday_weekday = 4 len([ ... date for date in ( ... begin_date + datetime.timedelta(days) ... for days in range((end_date - begin_date).days)) ... if calendar.weekday(date.year, date.month, date.day) = friday_weekday]) 9 -- \ “Selfish, adj. Devoid of consideration for the selfishness of | `\ others.” —Ambrose Bierce, _The Devil's Dictionary_, 1906 | _o__) | Ben Finney -- http://mail.python.org/mailman/listinfo/python-list
Re: weekdays in range
Ben Finney wrote: Jive Dadson notonthe...@noisp.com writes: Can someone think of an easy way to calculate the number of weekdays between two calendar dates (in Python)? That depends on what you mean by “weekdays”. import datetime begin_date = datetime.date(2009, 10, 9) end_date = datetime.date(2009, 10, 22) import calendar print calendar.month(2009, 10) October 2009 Mo Tu We Th Fr Sa Su 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 end_date - begin_date datetime.timedelta(13) If you're expecting to exclude Saturday and Sunday (i.e. if you expect the above result to be 9 days instead of 13), you can use other functions of the ‘calendar’ module; try starting with: friday_weekday = 4 len([ ... date for date in ( ... begin_date + datetime.timedelta(days) ... for days in range((end_date - begin_date).days)) ... if calendar.weekday(date.year, date.month, date.day) = friday_weekday]) 9 Thanks for your help. For a non-expert at Python, that last compound statement is pretty inscrutable. I am trying to scrute it. Wish me luck. -- http://mail.python.org/mailman/listinfo/python-list
Re: weekdays in range
Jive Dadson notonthe...@noisp.com writes: Ben Finney wrote: friday_weekday = 4 len([ ... date for date in ( ... begin_date + datetime.timedelta(days) ... for days in range((end_date - begin_date).days)) ... if calendar.weekday(date.year, date.month, date.day) = friday_weekday]) 9 Thanks for your help. For a non-expert at Python, that last compound statement is pretty inscrutable. I am trying to scrute it. Wish me luck. I'll help you by showing (liberally editing to make me look reptrospectively clever) how I built it up: import datetime import calendar import pprint begin_date = datetime.date(2009, 10, 9) end_date = datetime.date(2009, 10, 22) end_date - begin_date datetime.timedelta(13) (end_date - begin_date).days 13 range((end_date - begin_date).days) [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] begin_date + datetime.timedelta(0) datetime.date(2009, 10, 9) begin_date + datetime.timedelta(1) datetime.date(2009, 10, 10) (begin_date + datetime.timedelta(days) ... for days in range((end_date - begin_date).days)) generator object at 0xf78c08f0 [ ... date for date in ( ... begin_date + datetime.timedelta(days) ... for days in range((end_date - begin_date).days)) ... ] [datetime.date(2009, 10, 9), datetime.date(2009, 10, 10), datetime.date(2009, 10, 11), datetime.date(2009, 10, 12), datetime.date(2009, 10, 13), datetime.date(2009, 10, 14), datetime.date(2009, 10, 15), datetime.date(2009, 10, 16), datetime.date(2009, 10, 17), datetime.date(2009, 10, 18), datetime.date(2009, 10, 19), datetime.date(2009, 10, 20), datetime.date(2009, 10, 21)] pprint.pprint([ ... date for date in ( ... begin_date + datetime.timedelta(days) ... for days in range((end_date - begin_date).days)) ... ]) [datetime.date(2009, 10, 9), datetime.date(2009, 10, 10), datetime.date(2009, 10, 11), datetime.date(2009, 10, 12), datetime.date(2009, 10, 13), datetime.date(2009, 10, 14), datetime.date(2009, 10, 15), datetime.date(2009, 10, 16), datetime.date(2009, 10, 17), datetime.date(2009, 10, 18), datetime.date(2009, 10, 19), datetime.date(2009, 10, 20), datetime.date(2009, 10, 21)] len([ ... date for date in ( ... begin_date + datetime.timedelta(days) ... for days in range((end_date - begin_date).days)) ... ]) 13 friday_weekday = 4 calendar.weekday(begin_date.year, begin_date.month, begin_date.day) 4 calendar.weekday(end_date.year, end_date.month, end_date.day) 3 [ ... date for date in ( ... begin_date + datetime.timedelta(days) ... for days in range((end_date - begin_date).days)) ... if calendar.weekday(date.year, date.month, date.day) = friday_weekday] [datetime.date(2009, 10, 9), datetime.date(2009, 10, 12), datetime.date(2009, 10, 13), datetime.date(2009, 10, 14), datetime.date(2009, 10, 15), datetime.date(2009, 10, 16), datetime.date(2009, 10, 19), datetime.date(2009, 10, 20), datetime.date(2009, 10, 21)] pprint.pprint([ ... date for date in ( ... begin_date + datetime.timedelta(days) ... for days in range((end_date - begin_date).days)) ... if calendar.weekday(date.year, date.month, date.day) = friday_weekday]) [datetime.date(2009, 10, 9), datetime.date(2009, 10, 12), datetime.date(2009, 10, 13), datetime.date(2009, 10, 14), datetime.date(2009, 10, 15), datetime.date(2009, 10, 16), datetime.date(2009, 10, 19), datetime.date(2009, 10, 20), datetime.date(2009, 10, 21)] len([ ... date for date in ( ... begin_date + datetime.timedelta(days) ... for days in range((end_date - begin_date).days)) ... if calendar.weekday(date.year, date.month, date.day) = friday_weekday]) 9 -- \ “We are all agreed that your theory is crazy. The question that | `\ divides us is whether it is crazy enough to have a chance of | _o__)being correct.” —Niels Bohr (to Wolfgang Pauli), 1958 | Ben Finney -- http://mail.python.org/mailman/listinfo/python-list
Re: weekdays in range
Wow. It's a danged tutorial. Thanks again. Take a break. -- http://mail.python.org/mailman/listinfo/python-list
Re: weekdays in range
Jive Dadson wrote: div class=moz-text-flowed style=font-family: -moz-fixedWow. It's a danged tutorial. Thanks again. Take a break. /div Ben Finney's method is a very good approach, and an experienced Python programmer would consider it straightforward. But I have to ask whether the range of dates you might be considering could be large. For example, if you want to know how many week-days there are between a date in 1904 and 2207, it'd build a list of some 110 thousand items. There are other approaches which would be faster, consume much less memory, and be much harder to read. I'm not offering to debug it, but here's such an approach, subject to the old plus/minus one bugs. def epochweekdays (datetimeobj) Given an arbitrary Monday long ago, figure out how many weekdays have occurred between that day and the argument subtract to get total_days return int(totaldays/7) * 5 + max(totaldays%7, 5) Now, your answer is just epochweekdays(b) - epochweekdays(a) DaveA -- http://mail.python.org/mailman/listinfo/python-list
