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
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Python tracker <[email protected]>
<http://bugs.python.org/issue16810>
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