Why can't timedeltas be divided?
maric at aristote.info
Thu May 25 14:02:15 CEST 2006
oups ididn't post it to the good thread :)
Le Jeudi 25 Mai 2006 01:10, vous avez écrit :
> The ratio of two durations has no meaning???
Oh, sorry, sure it has, I wanted to say "it has no meaning in timedelta
It's a ratio (no dimension) not a duration. In that sense the expected result
should be a float, and the proposed operator will break the timedelta's
t, u, v <- timedeltas
t+u # valid
t / u # valid
t / u + v # invalid while all terms are valids
It's a big design flaw and I think it's the full answer to the original
Le Jeudi 25 Mai 2006 02:26, Robert Kern a écrit :
> > what you want is :
> > num_weeks = time_diff.days / 7
> > or
> > num_weeks = (time_diff / 7).days
> Uh, no. Besides the integer division problem in your first line, keep in
> mind that the .days attribute does not give you the time interval measured
> in days. It gives you the number of *whole* days in the interval. The first
> method will be incorrect if time_diff is not an even multiple of 1 day.
> The latter will be incorrect if time_diff is not an even multiple of 7 days.
In fact i was computing the exact number of whole weeks in the delta. In
respect of that both expression are perfectly correct, but the second one
isn't clear IMO (why this "days" attribute should give me the number of
This said it's not hard to figure out the correct expression of the decimal
value of weeks in deltas (discarding the microseconds which are not
num_weeks = (time_diff.days * 24* 3600 + time_diff.seconds) / (7.*24*3600)
If I need to do much of these in a piece of code I would probably define some
helper functions like this :
def tomicroseconds(td) :
return td.days * 24* 3600 * 10**6 +
td.seconds * 10 ** 6 + td.microseconds
def toseconds(td) : return float(tomicroseonds(td)) / 10 ** 6
tominute, tohours, todays, toweeks, etc...
and use float and int / and % operators.
This is an easy and clean implementation IMHO.
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