How do I convert datetime.timedelta to minutes, ho

Question

I get a start_date like this:

from django.utils.timezone import utc
import datetime

start_date = datetime.datetime.utcnow().replace(tzinfo=utc)
end_date = datetime.datetime.utcnow().replace(tzinfo=utc)
duration = end_date - start_date

I get output like this:

datetime.timedelta(0, 5, 41038)

How do I convert this into normal time like the following?

10 minutes, 1 hour like this

Answer 1

I defined own helper function to convert timedelta object to 'HH:MM:SS' format - only hours, minutes and seconds, without changing hours to days.

def format_timedelta(td):
    hours, remainder = divmod(td.total_seconds(), 3600)
    minutes, seconds = divmod(remainder, 60)
    hours, minutes, seconds = int(hours), int(minutes), int(seconds)
    if hours < 10:
        hours = '0%s' % int(hours)
    if minutes < 10:
        minutes = '0%s' % minutes
    if seconds < 10:
        seconds = '0%s' % seconds
    return '%s:%s:%s' % (hours, minutes, seconds)

Answer 2

A datetime.timedelta corresponds to the difference between two dates, not a date itself. It's only expressed in terms of days, seconds, and microseconds, since larger time units like months and years don't decompose cleanly (is 30 days 1 month or 0.9677 months?).

If you want to convert a timedelta into hours and minutes, you can use the total_seconds() method to get the total number of seconds and then do some math:

x = datetime.timedelta(1, 5, 41038)  # Interval of 1 day and 5.41038 seconds
secs = x.total_seconds()
hours = int(secs / 3600)
minutes = int(secs / 60) % 60

Answer 3

Just use strftime :)

Something like that:

my_date = datetime.datetime(2013, 1, 7, 10, 31, 34, 243366, tzinfo=<UTC>)
print(my_date.strftime("%Y, %d %B"))

---

After edited your question to format timedelta, you could use:

def timedelta_tuple(timedelta_object):
   return timedelta_object.days, timedelta_object.seconds//3600, (timedelta_object.seconds//60)%60

Answer 4

I don't think it's a good idea to caculate yourself.

If you just want a pretty output, just covert it into str with str() function or directly print() it.

And if there's further usage of the hours and minutes, you can parse it to datetime object use datetime.strptime()(and extract the time part with datetime.time() mehtod), for example:

import datetime

delta = datetime.timedelta(seconds=10000)
time_obj = datetime.datetime.strptime(str(delta),'%H:%M:%S').time()

Answer 5

There is no need for custom helper functions if all we need is to print the string of the form [D day[s], ][H]H:MM:SS[.UUUUUU]. timedelta object supports str() operation that will do this. It works even in Python 2.6.

>>> from datetime import timedelta
>>> timedelta(seconds=90136)
datetime.timedelta(1, 3736)
>>> str(timedelta(seconds=90136))
'1 day, 1:02:16'

Answer 6

Another alternative for this (older) question:

import datetime
import pytz
import time

pacific=pytz.timezone('US/Pacific')
now=datetime.datetime.now()
# pacific.dst(now).total_seconds() yields 3600 secs. [aka 1 hour]
time.strftime("%-H", time.gmtime(pacific.dst(now).total_seconds()))
'1'

The above is a good way to tell if your current time zone is actually in daylight savings time or not. (It provides an offset of 0 or 1.) Anyway, the real work is being done by time.strftime("%H:%M:%S", time.gmtime(36901)) which does work on the output of gmtime().

>>> time.strftime("%H:%M:%S",time.gmtime(36901))  # secs = 36901
'10:15:01'

And, that's it! (NOTE: Here's a link to format specifiers for time.strftime(). ...)