New submission from Glenn Linderman: Docs say:
date.timetuple() Return a time.struct_time such as returned by time.localtime(). The hours, minutes and seconds are 0, and the DST flag is -1. d.timetuple() is equivalent to time.struct_time((d.year, d.month, d.day, 0, 0, 0, d.weekday(), yday, -1)), where yday = d.toordinal() - date(d.year, 1, 1).toordinal() + 1 is the day number within the current year starting with 1 for January 1st. However, timetuple's 7th element has a range of 0-6 where 0 is Sunday, and d.weekday has a range of 0-6 where 0 is Monday. So the claim of equivalence is false. "d.weekday()" in the above could be replaced by "( d.weekday() + 1 ) % 7" I guess datetime consistently uses 0==Monday, and weeks starting on Monday, except for the timetuple (which probably has compatibility constraints which force it to return a different value, which I consider to be more correct). ---------- assignee: docs@python components: Documentation messages: 178477 nosy: docs@python, v+python priority: normal severity: normal status: open title: inconsistency in weekday versions: Python 3.3 _______________________________________ Python tracker <rep...@bugs.python.org> <http://bugs.python.org/issue16810> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: http://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com