[time-nuts] leapseconds, converting between GPS time (week, second) and UTC

[Mike Cook]

AFIK no programming systems professing POSIX compliance implements Leap Seconds. There is however a timezone UTCLeapSeconds will allow this apparently.

>> t1.TimeZone = ‘UTCLeapSeconds'

Obviously youR TZ data base needs to have it and your Python needs to be at the required level;

My test didn’t work, maybe I am not at the right levels.

>>> t1 = datetime
>>> t1.timezone = 'UTCLeapSeconds'
>>> t1 = datetime.datetime(2016,12,31,23,59,58)
>>> dt = datetime.timedelta(seconds=1)
>>> print t1
2016-12-31 23:59:58
>>> print t1+dt
2016-12-31 23:59:59
>>> print t1+dt+dt
2017-01-01 00:00:00
>>> print t1+dt+dt+dt
2017-01-01 00:00:01

I guess the same facility exists for other programming systems MATLAB has it I believe. Google is your friend.

Mike

> Le 15 janv. 2019 à 15:54, jimlux <jimlux at earthlink.net> a écrit :
> 
> I'm working with a variety of things which work in UTC or GPS week/millisecond, so we're doing a lot of conversion back and forth.
> (the spacecraft puts out time in week:millisecond, all the ground systems processing is in UTC)
> 
> The question comes up when converting back and forth, and whether various libraries handle leap seconds correctly.
> For now, I can use a hack of not computing back to 6 Jan 1980, but use an epoch like 15 Dec 2018 (week 2031: 518,400.000 seconds) and hope there's no leap second in the offing.
> 
> 
> For instance, in Python, if I do a datetime(2016,12,31,0,0,0) + timedelta(hours=30) does it come out as 1/1/2017 6:00:00 or 5:59:59  (it comes out 0600)
> 
> Similarly, does Excel's time formatting allow for some minutes having an extra second, or does it just assume all minutes are 60 seconds.
> 
> I'll probably test it for the cases I'm interested in (Ruby, Python, Excel, Matlab, Octave), but if someone else has already done it, then I've got something to cross check against.
> 
> 
> (python does NOT know about leap seconds)
> 
> import datetime
> 
> d = datetime.datetime(2016,12,31)
> 
> dt = datetime.timedelta(hours=30)
> 
> d
> Out[4]: datetime.datetime(2016, 12, 31, 0, 0)
> 
> dt
> Out[5]: datetime.timedelta(1, 21600)
> 
> d+dt
> Out[6]: datetime.datetime(2017, 1, 1, 6, 0)
> 
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