[Numpy-discussion] Making datetime64 timezone naive

Chris Barker - NOAA Federal chris.barker at noaa.gov
Mon Oct 19 20:54:49 EDT 2015

> This is fine.  Just be aware that *naive* datetimes will also have the PEP 495 "fold" attribute in Python 3.6.  You are free to ignore it, but you will loose the ability to round-trip between naive stdlib datetimes and numpy.datetime64.

Sigh. I can see why it's there ( primarily to support now(), I
suppose). But a naive datetime doesn't have a timezone, so how could
you know what  time one actually corresponds to if fold is True? And
what could you do with it if you did know?

I've always figured that if you are using naive time for times in a
timezone that has DST, than you'd better know wether you were in DST
or not.

(Which fold tells you, I guess) but the fold isn't guaranteed to be an
hour is it? So without more info, what can you do? And if the fold bit
is False, then you still have no idea if you are in DST or not.

And then what if you attach a timezone to it? Then the fold bit could
be wrong...

I take it back, I can't see why the fold bit could be anything but
confusing for a naive datetime. :-)

Anyway, all I can see to do here is for the datetime64 docs to say
that fold is ignored if it's there.

But what should datetime64 do when provided with a datetime with a timezone?

- Raise an exception?
- ignore the timezone?
- Convert to UTC?

If the time zone is ignored, then you could get DST and non DST times
in the same array - that could be ugly.

Is there any way to query a timezone object to ask if it's a constant-offset?

And yes, I did mean "most". There is no way I'm ever going to
introduce a three letter "timezone" abbreviation in one of these


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