Alternative to datetime.datetime.strftime() for cutting out unneeded units of time

Question

I'd like to compare two date times with an exact string output such as "3 hours, 2, minutes, 46 seconds". Now, say I have two datetime objects in UTC, I need them to be able to be an arbitrary amount but still cut off unneeded values. Example:

>>> import datetime
>>> date1 = datetime.datetime.now()
>>> date2 = datetime.datetime.fromtimestamp(123456789)
>>> date2 - date1
datetime.timedelta(-16362, 59566, 655021)

Now I could just use datetime.datetime.strftime() as a follow up but with the wide range of values I can encounter (1 second to months) I want to cut off the unneeded values and avoid "0 months, 0 weeks... etc." and "1 hour, 0 minutes, 0 seconds" without rounding (values are always positive). Is there an effient way to dynamically do this to only show relevent time values? I've looked through other questions similar to How do I find the time difference between two datetime objects in python? but none fully answer my question and I cannot figure it out on my own after many hours of tinkering.

Exactly which values are "relevant" and which are "unneeded"? For example, if you just want resolution down to the day, you can use the `datetime.date()` method to convert to a `date` object. — jwodder, Sep 16 '18 at 05:12
I'm almost certain that this would need to be done with your own function. datetime objects have attributes for `months`, etc. so you would need to implement these yourself. — IllustriousMagenta, Sep 16 '18 at 05:25
@jwodder If I am given a date which is "4 hours, 0 minutes, and 10 seconds" difference to the current time, I don't want to return the "0 minutes". Same goes for not wanting to return "0 days" but adding that if there *are* days. Basically I want to return all values which do not have a value equal to zero — Matthew, Sep 16 '18 at 16:45

score 1 · Accepted Answer · answered Sep 17 '18 at 19:26

Question: ... do this to only show relevent time values?

The following implements your own class Timedelta:

class Timedelta(object):
    def __init__(self, td):
        self._td = td

        def _calc(s, f):
            _s = int(s % f)
            return _s, int(s / f)

        s, self.hours = _calc(self._td.seconds, (60 * 60))
        self.seconds, self.minutes = _calc(s, 60)

    def __str__(self):
        r = ""
        for v in [('Days', self._td.days), ('Hours', self.hours), ('Minutes', self.minutes), ('Seconds', self.seconds)]:
            if v[1] != 0:
                lingua = v[0]
                if v[1] == 1: lingua = v[0][:-1]
                r += "{} {} ".format(v[1], lingua)
        return  r

    def __repr__(self):
        return "({}) Days:{} Hours:{} Minutes:{} Seconds:{}"\
            .format((self._td.seconds), self._td.days, self.hours, self.minutes, self.seconds)

Usage:

date1 = dt.datetime(2018, 9, 17, hour=18, minute=23, second=0)
date2 = dt.datetime(2018, 9, 17, hour=22, minute=23, second=10)
dtt = date2 - date1
print("dtt:{}".format(dtt))

tdelta = Timedelta(date2 - date1)
print("Timedelta::\__repr\__: {}".format(tdelta.__repr__()))
print("Timedelta::\__str\__: {}".format(tdelta))

Output:

dtt:4:00:10
Timedelta::__repr__: (14410) Days:0 Hours:4 Minutes:0 Seconds:10
Timedelta:__str__: 4 Hours 10 Seconds

Tested with Python:3.4.2

Alternative to datetime.datetime.strftime() for cutting out unneeded units of time

1 Answers1