[python] Convert a timedelta to days, hours and minutes

I've got a timedelta. I want the days, hours and minutes from that - either as a tuple or a dictionary... I'm not fussed.

I must have done this a dozen times in a dozen languages over the years but Python usually has a simple answer to everything so I thought I'd ask here before busting out some nauseatingly simple (yet verbose) mathematics.

Mr Fooz raises a good point.

I'm dealing with "listings" (a bit like ebay listings) where each one has a duration. I'm trying to find the time left by doing when_added + duration - now

Am I right in saying that wouldn't account for DST? If not, what's the simplest way to add/subtract an hour?

I used the following:

delta = timedelta()
totalMinute, second = divmod(delta.seconds, 60)
hour, minute = divmod(totalMinute, 60)
print(f"{hour}h{minute:02}m{second:02}s")

I found the easiest way is using str(timedelta). It will return a sting formatted like 3 days, 21:06:40.001000, and you can parse hours and minutes using simple string operations or regular expression.

timedeltas have a days and seconds attribute .. you can convert them yourself with ease.

This is a bit more compact, you get the hours, minutes and seconds in two lines.

days = td.days
hours, remainder = divmod(td.seconds, 3600)
minutes, seconds = divmod(remainder, 60)
# If you want to take into account fractions of a second
seconds += td.microseconds / 1e6

I don't understand

days, hours, minutes = td.days, td.seconds // 3600, td.seconds // 60 % 60

how about this

days, hours, minutes = td.days, td.seconds // 3600, td.seconds % 3600 / 60.0

You get minutes and seconds of a minute as a float.

Here is a little function I put together to do this right down to microseconds:

def tdToDict(td:datetime.timedelta) -> dict:
    def __t(t, n):
        if t < n: return (t, 0)
        v = t//n
        return (t -  (v * n), v)
    (s, h) = __t(td.seconds, 3600)
    (s, m) = __t(s, 60)    
    (micS, milS) = __t(td.microseconds, 1000)

    return {
         'days': td.days
        ,'hours': h
        ,'minutes': m
        ,'seconds': s
        ,'milliseconds': milS
        ,'microseconds': micS
    }

Here is a version that returns a tuple:

# usage: (_d, _h, _m, _s, _mils, _mics) = tdTuple(td)
def tdTuple(td:datetime.timedelta) -> tuple:
    def _t(t, n):
        if t < n: return (t, 0)
        v = t//n
        return (t -  (v * n), v)
    (s, h) = _t(td.seconds, 3600)
    (s, m) = _t(s, 60)    
    (mics, mils) = _t(td.microseconds, 1000)
    return (td.days, h, m, s, mics, mils)

days, hours, minutes = td.days, td.seconds // 3600, td.seconds // 60 % 60

As for DST, I think the best thing is to convert both datetime objects to seconds. This way the system calculates DST for you.

>>> m13 = datetime(2010, 3, 13, 8, 0, 0)  # 2010 March 13 8:00 AM
>>> m14 = datetime(2010, 3, 14, 8, 0, 0)  # DST starts on this day, in my time zone
>>> mktime(m14.timetuple()) - mktime(m13.timetuple())     # difference in seconds
82800.0
>>> _/3600                                                # convert to hours
23.0