Paul Ganssle <[email protected]> added the comment:
Hi dh4931 — this is the expected result, assuming that the offsets changed
between those two dates in your system local time.
The .timestamp() method returns an epoch time, which is the number of seconds
since 1970-01-01T00:00:00 UTC, and so it is inherently timezone-aware. In
Python 3, naïve datetimes went from being "unitless datetimes" to representing
"local datetimes", and in certain situations (like calling `.timestamp()`),
your system's time zone is used.
If you want something that gives the number of seconds that has elapsed between
two naïve datetimes on the calendar and ignoring any daylight saving time
transitions, subtract them directly to get a timedelta, then divide the result
by a timedelta representing 1 second, like so:
>>> (datetime.datetime(1986, 5, 4, 7, 13, 22) - datetime.datetime(1986, 5,
4, 0, 0, 0)) / datetime.timedelta(seconds=1)
26002.0
>>> (datetime.datetime(1986, 5, 2, 7, 13, 22) - datetime.datetime(1986, 5,
2, 0, 0, 0)) / datetime.timedelta(seconds=1)
26002.0
----------
resolution: -> not a bug
stage: -> resolved
type: -> behavior
_______________________________________
Python tracker <[email protected]>
<https://bugs.python.org/issue41321>
_______________________________________
_______________________________________________
Python-bugs-list mailing list
Unsubscribe:
https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
