On Thu, Jun 9, 2011 at 1:38 AM, Pierre GM <pgmdevl...@gmail.com> wrote:
> > On Jun 9, 2011, at 2:22 AM, Mark Wiebe wrote: > > > > >>> np.array(['2011-03-12T13', '2012'], dtype='M8') > > > > array(['2011-03-12T13:00:00.000000-0600', > '2011-12-31T18:00:00.000000-0600'], dtype='datetime64[us]') > > > > > > Why is the second one not '2012-01-01T00:00:00-0600' ? > > > > > > This is because dates are stored at midnight UTC, and when converted to > local time for the default time-based printing, that changes slightly. > > > ISO8601 specifies to interpret an input in local time if no "Z" or > timezone offset is given, so that's why the first one matches. I haven't > been able to think of a way around it other than putting warnings in the > documentation, and have made 'today' and 'now' throw errors if you try to > use them as times or dates respectively. > > > > I see the logic, but I don't like it at all. I would expect the date to > be stored in the local time zone by default (that is, if no other time zone > info is available). > > > > It's not satisfying to me either, but I haven't been able to think of a > solution I like. The idea of converting from 'D' to 's' metadata depending > on the timezone setting of your computer feels worse to me than the current > approach, but if someone has an idea that's better I'm all ears. > > A simpler approach could be to drop ISO8601 recommendation and assume that > any single datetime is expressed in UTC by default. Managing time zones > would then be let to some specific subclass... Doing nothing with time zones would mean this prints by default: >>> np.datetime_as_string(np.datetime64('now')) '2011-06-09T20:03:20Z' instead of this: >>> np.datetime64('now') numpy.datetime64('2011-06-09T15:03:28-0500','s') The issue above is the relationship between dates and datetimes, which the NumPy datetime64 has in a continuum. From a UTC perspective, this is very natural, but it does cause a few hiccups. I'll make a thread at some point for discussing the time zone issues. -Mark
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