python – How do I convert seconds to hours, minutes and seconds?

The Question :

520 people think this question is useful

I have a function that returns information in seconds, but I need to store that information in hours:minutes:seconds.

Is there an easy way to convert the seconds to this format in Python?

The Question Comments :

The Answer 1

857 people think this answer is useful

You can use datetime.timedelta function:

>>> import datetime
>>> str(datetime.timedelta(seconds=666))
'0:11:06'

The Answer 2

639 people think this answer is useful

By using the divmod() function, which does only a single division to produce both the quotient and the remainder, you can have the result very quickly with only two mathematical operations:

m, s = divmod(seconds, 60)
h, m = divmod(m, 60)

And then use string formatting to convert the result into your desired output:

print('{:d}:{:02d}:{:02d}'.format(h, m, s)) # Python 3
print(f'{h:d}:{m:02d}:{s:02d}') # Python 3.6+

The Answer 3

78 people think this answer is useful

I can hardly name that an easy way (at least I can’t remember the syntax), but it is possible to use time.strftime, which gives more control over formatting:

from time import strftime
from time import gmtime

strftime("%H:%M:%S", gmtime(666))
'00:11:06'

strftime("%H:%M:%S", gmtime(60*60*24))
'00:00:00'

gmtime is used to convert seconds to special tuple format that strftime() requires.

Note: Truncates after 23:59:59

The Answer 4

46 people think this answer is useful

Using datetime:

With the ':0>8' format:

from datetime import timedelta

"{:0>8}".format(str(timedelta(seconds=66)))
# Result: '00:01:06'

"{:0>8}".format(str(timedelta(seconds=666777)))
# Result: '7 days, 17:12:57'

"{:0>8}".format(str(timedelta(seconds=60*60*49+109)))
# Result: '2 days, 1:01:49'

Without the ':0>8' format:

"{}".format(str(timedelta(seconds=66)))
# Result: '00:01:06'

"{}".format(str(timedelta(seconds=666777)))
# Result: '7 days, 17:12:57'

"{}".format(str(timedelta(seconds=60*60*49+109)))
# Result: '2 days, 1:01:49'

Using time:

from time import gmtime
from time import strftime

# NOTE: The following resets if it goes over 23:59:59!

strftime("%H:%M:%S", gmtime(125))
# Result: '00:02:05'

strftime("%H:%M:%S", gmtime(60*60*24-1))
# Result: '23:59:59'

strftime("%H:%M:%S", gmtime(60*60*24))
# Result: '00:00:00'

strftime("%H:%M:%S", gmtime(666777))
# Result: '17:12:57'
# Wrong

The Answer 5

25 people think this answer is useful

This is my quick trick:

from humanfriendly import format_timespan
secondsPassed = 1302
format_timespan(secondsPassed)
# '21 minutes and 42 seconds'

For more info Visit: https://humanfriendly.readthedocs.io/en/latest/#humanfriendly.format_timespan

The Answer 6

10 people think this answer is useful

If you need to get datetime.time value, you can use this trick:

my_time = (datetime(1970,1,1) + timedelta(seconds=my_seconds)).time()

You cannot add timedelta to time, but can add it to datetime.

UPD: Yet another variation of the same technique:

my_time = (datetime.fromordinal(1) + timedelta(seconds=my_seconds)).time()

Instead of 1 you can use any number greater than 0. Here we use the fact that datetime.fromordinal will always return datetime object with time component being zero.

The Answer 7

9 people think this answer is useful

This is how I got it.

def sec2time(sec, n_msec=3):
    ''' Convert seconds to 'D days, HH:MM:SS.FFF' '''
    if hasattr(sec,'__len__'):
        return [sec2time(s) for s in sec]
    m, s = divmod(sec, 60)
    h, m = divmod(m, 60)
    d, h = divmod(h, 24)
    if n_msec > 0:
        pattern = '%%02d:%%02d:%%0%d.%df' % (n_msec+3, n_msec)
    else:
        pattern = r'%02d:%02d:%02d'
    if d == 0:
        return pattern % (h, m, s)
    return ('%d days, ' + pattern) % (d, h, m, s)

Some examples:

$ sec2time(10, 3)
Out: '00:00:10.000'

$ sec2time(1234567.8910, 0)
Out: '14 days, 06:56:07'

$ sec2time(1234567.8910, 4)
Out: '14 days, 06:56:07.8910'

$ sec2time([12, 345678.9], 3)
Out: ['00:00:12.000', '4 days, 00:01:18.900']

The Answer 8

8 people think this answer is useful

The following set worked for me.

def sec_to_hours(seconds):
    a=str(seconds//3600)
    b=str((seconds%3600)//60)
    c=str((seconds%3600)%60)
    d=["{} hours {} mins {} seconds".format(a, b, c)]
    return d


print(sec_to_hours(10000))
# ['2 hours 46 mins 40 seconds']

print(sec_to_hours(60*60*24+105))
# ['24 hours 1 mins 45 seconds']

The Answer 9

5 people think this answer is useful

hours (h) calculated by floor division (by //) of seconds by 3600 (60 min/hr * 60 sec/min)

minutes (m) calculated by floor division of remaining seconds (remainder from hour calculation, by %) by 60 (60 sec/min)

similarly, seconds (s) by remainder of hour and minutes calculation.

Rest is just string formatting!

def hms(seconds):
    h = seconds // 3600
    m = seconds % 3600 // 60
    s = seconds % 3600 % 60
    return '{:02d}:{:02d}:{:02d}'.format(h, m, s)

print(hms(7500))  # Should print 02h05m00s

The Answer 10

4 people think this answer is useful

dateutil.relativedelta is convenient if you need to access hours, minutes and seconds as floats as well. datetime.timedelta does not provide a similar interface.

from dateutil.relativedelta import relativedelta
rt = relativedelta(seconds=5440)
print(rt.seconds)
print('{:02d}:{:02d}:{:02d}'.format(
    int(rt.hours), int(rt.minutes), int(rt.seconds)))

Prints

40.0
01:30:40

The Answer 11

0 people think this answer is useful

In my case I wanted to achieve format “HH:MM:SS.fff”. I solved it like this:

timestamp = 28.97000002861023
str(datetime.fromtimestamp(timestamp)+timedelta(hours=-1)).split(' ')[1][:12]
'00:00:28.970'

The Answer 12

-2 people think this answer is useful

You can divide seconds by 60 to get the minutes

import time
seconds = time.time()
minutes = seconds / 60
print(minutes)

When you divide it by 60 again, you will get the hours

The Answer 13

-4 people think this answer is useful

division = 3623 // 3600 #to hours
division2 = 600 // 60 #to minutes
print (division) #write hours
print (division2) #write minutes

PS My code is unprofessional

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