How do I convert datetime.timedelta to minutes, hours in Python?

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

possible duplicate of [Convert a timedelta to days, hours and minutes](http://stackoverflow.com/questions/2119472/convert-a-timedelta-to-days-hours-and-minutes) — Anto, Jan 06 '14 at 15:18
The fact that there are so many people looking this up indicates that the datetime module should have this basic feature... — theQuestionMan, Feb 16 '21 at 04:14

abarnert · Accepted Answer · 2013-01-07T05:50:19.527

There's no built-in formatter for timedelta objects, but it's pretty easy to do it yourself:

days, seconds = duration.days, duration.seconds
hours = days * 24 + seconds // 3600
minutes = (seconds % 3600) // 60
seconds = seconds % 60

Or, equivalently, if you're in Python 2.7+ or 3.2+:

seconds = duration.total_seconds()
hours = seconds // 3600
minutes = (seconds % 3600) // 60
seconds = seconds % 60

Now you can print it however you want:

'{} minutes, {} hours'.format(minutes, hours)

For example:

def convert_timedelta(duration):
    days, seconds = duration.days, duration.seconds
    hours = days * 24 + seconds // 3600
    minutes = (seconds % 3600) // 60
    seconds = (seconds % 60)
    return hours, minutes, seconds
td = datetime.timedelta(2, 7743, 12345)
hours, minutes, seconds = convert_timedelta(td)
print '{} minutes, {} hours'.format(minutes, hours)

This will print:

9 minutes, 50 hours

If you want to get "10 minutes, 1 hour" instead of "10 minutes, 1 hours", you need to do that manually too:

print '{} minute{}, {} hour{}'.format(minutes, 's' if minutes != 1 else '',
                                      hours, 's' if minutes != 1 else '')

Or you may want to write an english_plural function to do the 's' bits for you, instead of repeating yourself.

From your comments, it sounds like you actually want to keep the days separate. That's even easier:

def convert_timedelta(duration):
    days, seconds = duration.days, duration.seconds
    hours = seconds // 3600
    minutes = (seconds % 3600) // 60
    seconds = (seconds % 60)
    return days, hours, minutes, seconds

If you want to convert this to a single value to store in a database, then convert that single value back to format it, do this:

def dhms_to_seconds(days, hours, minutes, seconds):
    return (((days * 24) + hours) * 60 + minutes) * 60 + seconds

def seconds_to_dhms(seconds):
    days = seconds // (3600 * 24)
    hours = (seconds // 3600) % 24
    minutes = (seconds // 60) % 60
    seconds = seconds % 60
    return days, hours, minutes, seconds

So, putting it together:

def store_timedelta_in_database(thingy, duration):
    seconds = dhms_to_seconds(*convert_timedelta(duration))
    db.execute('INSERT INTO foo (thingy, duration) VALUES (?, ?)',
               thingy, seconds)
    db.commit()

def print_timedelta_from_database(thingy):
    cur = db.execute('SELECT duration FROM foo WHERE thingy = ?', thingy)
    seconds = int(cur.fetchone()[0])
    days, hours, minutes, seconds = seconds_to_dhms(seconds)
    print '{} took {} minutes, {} hours, {} days'.format(thingy, minutes, hours, days)

Also, What if the duration exceeds more than a day? I want to show it in days format. — user1881957, Jan 07 '13 at 05:29
One more problem, since the hours and minutes are store in a separate variable how can I insert all those into a single coloumn called attribute. — user1881957, Jan 07 '13 at 05:33
If you want to print days, that makes it even easier; just don't mix them in. I'll edit the answer to demonstrate. As for the second question, I'm not sure what you mean. What format is `attribute` supposed to be? My first guess is that you actually want to store just a single `seconds` number in the database, and parse it each time you retrieve it; is that what you mean? — abarnert, Jan 07 '13 at 05:43
Edit the question with codes. Yes, I will store in seconds in database. Then while showing to the user I want to parse as suitable in minutes, hours or days. Thanks — user1881957, Jan 07 '13 at 05:51
@user1881957: I'm not sure what you mean by "Edit the question with codes." You haven't edited the question. I already edited the answer to show most likely all the conversion functions you'll ever need (and, if not, it should be simple to write your own using them as a guide). Is there something else you need? — abarnert, Jan 07 '13 at 06:15
Opps Sorry I meant edit the answers with codes. All set now. Thanks — user1881957, Jan 07 '13 at 08:21
Hi Sorry to bother you again. I used your seconds_to_dhms function I get duration like (0, 1, 2, 4) {days, hours, minutes and seconds} respectively. I am storing duration like this in database. Now to show this to the user in Django template in a proper way how can I do this? — user1881957, Jan 07 '13 at 09:08
There IS a built-in formatter for ```timedelta``` objects. Have a look at my answer. — wombatonfire, May 14 '17 at 14:35

score 25 · Answer 2 · answered Jan 07 '13 at 05:04

25

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

answered Jan 07 '13 at 05:04

Adam Rosenfield

375,615
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501
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AttributeError: 'datetime.timedelta' object has no attribute 'total_seconds' – user1881957 Jan 07 '13 at 05:28
1

@user1881957: `total_seconds` was added in… I believe 2.7 and 3.2. If you're using 2.6 or 3.1 or earlier, use the explicit `days`/`seconds` alternative from my answer; it's not much different. – abarnert Jan 07 '13 at 05:42

score 20 · Answer 3 · answered May 14 '17 at 14:33

20

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'

answered May 14 '17 at 14:33

wombatonfire

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23
35

2

Except for the weird output format requirement of the question (which I think is not really the real requirement, just the OP having problems expressing themselves), this should be the accepted answer. – Peter Mortensen Oct 29 '18 at 23:22

score 7 · Answer 4 · answered Feb 16 '19 at 08:59

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()

Yam Mesicka · Answer 5 · 2013-01-07T05:03:51.590

4

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

edited Jan 07 '13 at 05:03

answered Jan 07 '13 at 04:55

Yam Mesicka

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score 3 · Answer 6 · answered Mar 03 '16 at 08:29

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)

ingyhere · Answer 7 · 2021-06-14T20:14:07.247

An alternative for this (older) question is to create a relative time from a timedelta converted to seconds. This can be accomplished using the time.gmtime(...) method that accepts seconds since the epoch:

>>> 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() so the difference can be truncated to any units, as needed. ...)

Notably, this technique is also a great way to tell if your current time zone is actually in daylight savings time or not. (It provides an offset of 0 or 1 hours meaning it can be interpreted basically as a boolean.)

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'

This can be rendered to a method is_standard_time(...) where 1 means true and 0 means false.

score 2 · Answer 8 · answered May 19 '20 at 04:54

# Try this code
from datetime import timedelta

class TimeDelta(timedelta):
    def __str__(self):
        _times = super(TimeDelta, self).__str__().split(':')
        if "," in _times[0]:
            _hour = int(_times[0].split(',')[-1].strip())
            if _hour:
                _times[0] += " hours" if _hour > 1 else " hour"
            else:
                _times[0] = _times[0].split(',')[0]
        else:
            _hour = int(_times[0].strip())
            if _hour:
                _times[0] += " hours" if _hour > 1 else " hour"
            else:
                _times[0] = ""
        _min = int(_times[1])
        if _min:
            _times[1] += " minutes" if _min > 1 else " minute"
        else:
            _times[1] = ""
        _sec = int(_times[2])
        if _sec:
            _times[2] += " seconds" if _sec > 1 else " second"
        else:
            _times[2] = ""
        return ", ".join([i for i in _times if i]).strip(" ,").title()

# Test
>>> str(TimeDelta(seconds=10))
'10 Seconds'
>>> str(TimeDelta(seconds=60))
'01 Minute'
>>> str(TimeDelta(seconds=90))
'01 Minute, 30 Seconds'
>>> str(TimeDelta(seconds=3000))
'50 Minutes'
>>> str(TimeDelta(seconds=3600))
'1 Hour'
>>> str(TimeDelta(seconds=3690))
'1 Hour, 01 Minute, 30 Seconds'
>>> str(TimeDelta(seconds=3660))
'1 Hour, 01 Minute'
>>> str(TimeDelta(seconds=3630))
'1 Hour, 30 Seconds'
>>> str(TimeDelta(seconds=3600*20))
'20 Hours'
>>> str(TimeDelta(seconds=3600*20 + 3000))
'20 Hours, 50 Minutes'
>>> str(TimeDelta(seconds=3600*20 + 3630))
'21 Hours, 30 Seconds'
>>> str(TimeDelta(seconds=3600*20 + 3660))
'21 Hours, 01 Minute'
>>> str(TimeDelta(seconds=3600*20 + 3690))
'21 Hours, 01 Minute, 30 Seconds'
>>> str(TimeDelta(seconds=3600*24))
'1 Day'
>>> str(TimeDelta(seconds=3600*24 + 10))
'1 Day, 10 Seconds'
>>> str(TimeDelta(seconds=3600*24 + 60))
'1 Day, 01 Minute'
>>> str(TimeDelta(seconds=3600*24 + 90))
'1 Day, 01 Minute, 30 Seconds'
>>> str(TimeDelta(seconds=3600*24 + 3000))
'1 Day, 50 Minutes'
>>> str(TimeDelta(seconds=3600*24 + 3600))
'1 Day, 1 Hour'
>>> str(TimeDelta(seconds=3600*24 + 3630))
'1 Day, 1 Hour, 30 Seconds'
>>> str(TimeDelta(seconds=3600*24 + 3660))
'1 Day, 1 Hour, 01 Minute'
>>> str(TimeDelta(seconds=3600*24 + 3690))
'1 Day, 1 Hour, 01 Minute, 30 Seconds'
>>> str(TimeDelta(seconds=3600*24*2))
'2 Days'
>>> str(TimeDelta(seconds=3600*24*2 + 9999))
'2 Days, 2 Hours, 46 Minutes, 39 Seconds'

score 1 · Answer 9 · edited Oct 29 '18 at 23:20

1

Do you want to print the date in that format? This is the Python documentation: http://docs.python.org/2/library/datetime.html#strftime-strptime-behavior

>>> a = datetime.datetime(2013, 1, 7, 10, 31, 34, 243366)
>>> print a.strftime('%Y %d %B, %M:%S%p')
>>> 2013 07 January, 31:34AM

For the timedelta:

>>> a =  datetime.timedelta(0,5,41038)
>>> print '%s seconds, %s microseconds' % (a.seconds, a.microseconds)

But please notice, you should make sure it has the related value. For the above cases, it doesn't have the hours and minute values, and you should calculate from the seconds.

edited Oct 29 '18 at 23:20

Peter Mortensen

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answered Jan 07 '13 at 04:55

jinghli

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Thanks @abarnert. Looks he updated his questions. I also updated the answer. – jinghli Jan 07 '13 at 05:09

score -1 · Answer 10 · answered Oct 15 '15 at 01:22

-1

datetime.timedelta(hours=1, minutes=10)
#python 2.7

answered Oct 15 '15 at 01:22

user5435345

25
1

How do I convert datetime.timedelta to minutes, hours in Python?

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