Format timedelta to string


I’m having trouble formatting a datetime.timedelta object.

Here’s what I’m trying to do:
I have a list of objects and one of the members of the class of the object is a timedelta object that shows the duration of an event. I would like to display that duration in the format of hours:minutes.

I have tried a variety of methods for doing this and I’m having difficulty. My current approach is to add methods to the class for my objects that return hours and minutes. I can get the hours by dividing the timedelta.seconds by 3600 and rounding it. I’m having trouble with getting the remainder seconds and converting that to minutes.

By the way, I’m using Google AppEngine with Django Templates for presentation.

Asked By: mawcs



>>> str(datetime.timedelta(hours=10.56))

>>> td = datetime.timedelta(hours=10.505) # any timedelta object
>>> ':'.join(str(td).split(':')[:2])

Passing the timedelta object to the str() function calls the same formatting code used if we simply type print td. Since you don’t want the seconds, we can split the string by colons (3 parts) and put it back together with only the first 2 parts.

Answered By: joeforker

Following Joe’s example value above, I’d use the modulus arithmetic operator, thusly:

td = datetime.timedelta(hours=10.56)
td_str = "%d:%d" % (td.seconds/3600, td.seconds%3600/60)

Note that integer division in Python rounds down by default; if you want to be more explicit, use math.floor() or math.ceil() as appropriate.

Answered By: UltraNurd

You can just convert the timedelta to a string with str(). Here’s an example:

import datetime
start = datetime.datetime(2009,2,10,14,00)
end   = datetime.datetime(2009,2,10,16,00)
delta = end-start
# prints 2:00:00
Answered By: Parand

Thanks everyone for your help. I took many of your ideas and put them together, let me know what you think.

I added two methods to the class like this:

def hours(self):
    retval = ""
    if self.totalTime:
        hoursfloat = self.totalTime.seconds / 3600
        retval = round(hoursfloat)
    return retval

def minutes(self):
    retval = ""
    if self.totalTime:
        minutesfloat = self.totalTime.seconds / 60
        hoursAsMinutes = self.hours() * 60
        retval = round(minutesfloat - hoursAsMinutes)
    return retval

In my django I used this (sum is the object and it is in a dictionary):

<td>{{ sum.0 }}</td>    
<td>{{ sum.1.hours|stringformat:"d" }}:{{ sum.1.minutes|stringformat:"#02.0d" }}</td>
Answered By: mawcs

As you know, you can get the total_seconds from a timedelta object by accessing the .seconds attribute.

Python provides the builtin function divmod() which allows for:

s = 13420
hours, remainder = divmod(s, 3600)
minutes, seconds = divmod(remainder, 60)
print '{:02}:{:02}:{:02}'.format(int(hours), int(minutes), int(seconds))
# result: 03:43:40

or you can convert to hours and remainder by using a combination of modulo and subtraction:

# arbitrary number of seconds
s = 13420
# hours
hours = s // 3600 
# remaining seconds
s = s - (hours * 3600)
# minutes
minutes = s // 60
# remaining seconds
seconds = s - (minutes * 60)
# total time
print '{:02}:{:02}:{:02}'.format(int(hours), int(minutes), int(seconds))
# result: 03:43:40
Answered By: John Fouhy

My datetime.timedelta objects went greater than a day. So here is a further problem. All the discussion above assumes less than a day. A timedelta is actually a tuple of days, seconds and microseconds. The above discussion should use td.seconds as joe did, but if you have days it is NOT included in the seconds value.

I am getting a span of time between 2 datetimes and printing days and hours.

span = currentdt - previousdt
print '%d,%dn' % (span.days,span.seconds/3600)
Answered By: user198910
def td_format(td_object):
    seconds = int(td_object.total_seconds())
    periods = [
        ('year',        60*60*24*365),
        ('month',       60*60*24*30),
        ('day',         60*60*24),
        ('hour',        60*60),
        ('minute',      60),
        ('second',      1)

    for period_name, period_seconds in periods:
        if seconds > period_seconds:
            period_value , seconds = divmod(seconds, period_seconds)
            has_s = 's' if period_value > 1 else ''
            strings.append("%s %s%s" % (period_value, period_name, has_s))

    return ", ".join(strings)
Answered By: Adam Jacob Muller

Questioner wants a nicer format than the typical:

  >>> import datetime
  >>> datetime.timedelta(seconds=41000)
  datetime.timedelta(0, 41000)
  >>> str(datetime.timedelta(seconds=41000))
  >>> str(datetime.timedelta(seconds=4102.33))
  >>> str(datetime.timedelta(seconds=413302.33))
  '4 days, 18:48:22.330000'

So, really there’s two formats, one where days are 0 and it’s left out, and another where there’s text “n days, h:m:s”. But, the seconds may have fractions, and there’s no leading zeroes in the printouts, so columns are messy.

Here’s my routine, if you like it:

def printNiceTimeDelta(stime, etime):
    delay = datetime.timedelta(seconds=(etime - stime))
    if (delay.days > 0):
        out = str(delay).replace(" days, ", ":")
        out = "0:" + str(delay)
    outAr = out.split(':')
    outAr = ["%02d" % (int(float(x))) for x in outAr]
    out   = ":".join(outAr)
    return out

this returns output as dd:hh:mm:ss format:


I did think about adding years to this, but this is left as an exercise for the reader, since the output is safe at over 1 year:

>>> str(datetime.timedelta(seconds=99999999))
'1157 days, 9:46:39'
Answered By: Kevin J. Rice

He already has a timedelta object so why not use its built-in method total_seconds() to convert it to seconds, then use divmod() to get hours and minutes?

hours, remainder = divmod(myTimeDelta.total_seconds(), 3600)
minutes, seconds = divmod(remainder, 60)

# Formatted only for hours and minutes as requested
print '%s:%s' % (hours, minutes)

This works regardless if the time delta has even days or years.

Answered By: Bill Kidd
t1 = datetime.datetime.strptime(StartTime, "%H:%M:%S %d-%m-%y")

t2 = datetime.datetime.strptime(EndTime, "%H:%M:%S %d-%m-%y")

return str(t2-t1)

So for:

StartTime = '15:28:53 21-07-13'
EndTime = '15:32:40 21-07-13'


Answered By: Aviad
def seconds_to_time_left_string(total_seconds):
    s = int(total_seconds)
    years = s // 31104000
    if years > 1:
        return '%d years' % years
    s = s - (years * 31104000)
    months = s // 2592000
    if years == 1:
        r = 'one year'
        if months > 0:
            r += ' and %d months' % months
        return r
    if months > 1:
        return '%d months' % months
    s = s - (months * 2592000)
    days = s // 86400
    if months == 1:
        r = 'one month'
        if days > 0:
            r += ' and %d days' % days
        return r
    if days > 1:
        return '%d days' % days
    s = s - (days * 86400)
    hours = s // 3600
    if days == 1:
        r = 'one day'
        if hours > 0:
            r += ' and %d hours' % hours
        return r 
    s = s - (hours * 3600)
    minutes = s // 60
    seconds = s - (minutes * 60)
    if hours >= 6:
        return '%d hours' % hours
    if hours >= 1:
        r = '%d hours' % hours
        if hours == 1:
            r = 'one hour'
        if minutes > 0:
            r += ' and %d minutes' % minutes
        return r
    if minutes == 1:
        r = 'one minute'
        if seconds > 0:
            r += ' and %d seconds' % seconds
        return r
    if minutes == 0:
        return '%d seconds' % seconds
    if seconds == 0:
        return '%d minutes' % minutes
    return '%d minutes and %d seconds' % (minutes, seconds)

for i in range(10):
    print pow(8, i), seconds_to_time_left_string(pow(8, i))

1 1 seconds
8 8 seconds
64 one minute and 4 seconds
512 8 minutes and 32 seconds
4096 one hour and 8 minutes
32768 9 hours
262144 3 days
2097152 24 days
16777216 6 months
134217728 4 years
Answered By: Veselin Penev

I had a similar problem with the output of overtime calculation at work. The value should always show up in HH:MM, even when it is greater than one day and the value can get negative. I combined some of the shown solutions and maybe someone else find this solution useful. I realized that if the timedelta value is negative most of the shown solutions with the divmod method doesn’t work out of the box:

def td2HHMMstr(td):
  '''Convert timedelta objects to a HH:MM string with (+/-) sign'''
  if td < datetime.timedelta(seconds=0):
    td = -td
    sign = ''
  tdhours, rem = divmod(td.total_seconds(), 3600)
  tdminutes, rem = divmod(rem, 60)
  tdstr = '{}{:}:{:02d}'.format(sign, int(tdhours), int(tdminutes))
  return tdstr

timedelta to HH:MM string:

td2HHMMstr(datetime.timedelta(hours=1, minutes=45))

td2HHMMstr(datetime.timedelta(days=2, hours=3, minutes=2))

td2HHMMstr(datetime.timedelta(hours=-3, minutes=-2))

td2HHMMstr(datetime.timedelta(days=-35, hours=-3, minutes=-2))
Answered By: nealtz

I know that this is an old answered question, but I use datetime.utcfromtimestamp() for this. It takes the number of seconds and returns a datetime that can be formatted like any other datetime.

duration = datetime.utcfromtimestamp(end - begin)
print duration.strftime('%H:%M')

As long as you stay in the legal ranges for the time parts this should work, i.e. it doesn’t return 1234:35 as hours are <= 23.

Answered By: sjngm

I would seriously consider the Occam’s Razor approach here:

td = str(timedelta).split('.')[0]

This returns a string without the microseconds

If you want to regenerate the datetime.timedelta object, just do this:

h,m,s = re.split(':', td)
new_delta = datetime.timedelta(hours=int(h),minutes=int(m),seconds=int(s))

2 years in, I love this language!

Answered By: Pyrotherm

Please check this function – it converts timedelta object into string ‘HH:MM:SS’

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)
Answered By:
from django.utils.translation import ngettext

def localize_timedelta(delta):
    ret = []
    num_years = int(delta.days / 365)
    if num_years > 0:
        delta -= timedelta(days=num_years * 365)
        ret.append(ngettext('%d year', '%d years', num_years) % num_years)

    if delta.days > 0:
        ret.append(ngettext('%d day', '%d days', delta.days) % delta.days)

    num_hours = int(delta.seconds / 3600)
    if num_hours > 0:
        delta -= timedelta(hours=num_hours)
        ret.append(ngettext('%d hour', '%d hours', num_hours) % num_hours)

    num_minutes = int(delta.seconds / 60)
    if num_minutes > 0:
        ret.append(ngettext('%d minute', '%d minutes', num_minutes) % num_minutes)

    return ' '.join(ret)

This will produce:

>>> from datetime import timedelta
>>> localize_timedelta(timedelta(days=3660, minutes=500))
'10 years 10 days 8 hours 20 minutes'
Answered By: Artem Vasilev

I personally use the humanize library for this:

>>> import datetime
>>> humanize.naturalday(
>>> humanize.naturalday( - datetime.timedelta(days=1))
>>> humanize.naturalday(, 6, 5))
'Jun 05'
>>> humanize.naturaldate(, 6, 5))
'Jun 05 2007'
>>> humanize.naturaltime( - datetime.timedelta(seconds=1))
'a second ago'
>>> humanize.naturaltime( - datetime.timedelta(seconds=3600))
'an hour ago'

Of course, it doesn’t give you exactly the answer you were looking for (which is, indeed, str(timeA - timeB), but I have found that once you go beyond a few hours, the display becomes quickly unreadable. humanize has support for much larger values that are human-readable, and is also well localized.

It’s inspired by Django’s contrib.humanize module, apparently, so since you are using Django, you should probably use that.

Answered By: anarcat

Here is a general purpose function for converting either a timedelta object or a regular number (in the form of seconds or minutes, etc.) to a nicely formatted string. I took mpounsett’s fantastic answer on a duplicate question, made it a bit more flexible, improved readibility, and added documentation.

You will find that it is the most flexible answer here so far since it allows you to:

  1. Customize the string format on the fly instead of it being hard-coded.
  2. Leave out certain time intervals without a problem (see examples below).


from string import Formatter
from datetime import timedelta

def strfdelta(tdelta, fmt='{D:02}d {H:02}h {M:02}m {S:02}s', inputtype='timedelta'):
    """Convert a datetime.timedelta object or a regular number to a custom-
    formatted string, just like the stftime() method does for datetime.datetime

    The fmt argument allows custom formatting to be specified.  Fields can 
    include seconds, minutes, hours, days, and weeks.  Each field is optional.

    Some examples:
        '{D:02}d {H:02}h {M:02}m {S:02}s' --> '05d 08h 04m 02s' (default)
        '{W}w {D}d {H}:{M:02}:{S:02}'     --> '4w 5d 8:04:02'
        '{D:2}d {H:2}:{M:02}:{S:02}'      --> ' 5d  8:04:02'
        '{H}h {S}s'                       --> '72h 800s'

    The inputtype argument allows tdelta to be a regular number instead of the  
    default, which is a datetime.timedelta object.  Valid inputtype strings: 
        's', 'seconds', 
        'm', 'minutes', 
        'h', 'hours', 
        'd', 'days', 
        'w', 'weeks'

    # Convert tdelta to integer seconds.
    if inputtype == 'timedelta':
        remainder = int(tdelta.total_seconds())
    elif inputtype in ['s', 'seconds']:
        remainder = int(tdelta)
    elif inputtype in ['m', 'minutes']:
        remainder = int(tdelta)*60
    elif inputtype in ['h', 'hours']:
        remainder = int(tdelta)*3600
    elif inputtype in ['d', 'days']:
        remainder = int(tdelta)*86400
    elif inputtype in ['w', 'weeks']:
        remainder = int(tdelta)*604800

    f = Formatter()
    desired_fields = [field_tuple[1] for field_tuple in f.parse(fmt)]
    possible_fields = ('W', 'D', 'H', 'M', 'S')
    constants = {'W': 604800, 'D': 86400, 'H': 3600, 'M': 60, 'S': 1}
    values = {}
    for field in possible_fields:
        if field in desired_fields and field in constants:
            values[field], remainder = divmod(remainder, constants[field])
    return f.format(fmt, **values)


>>> td = timedelta(days=2, hours=3, minutes=5, seconds=8, microseconds=340)

>>> print strfdelta(td)
02d 03h 05m 08s

>>> print strfdelta(td, '{D}d {H}:{M:02}:{S:02}')
2d 3:05:08

>>> print strfdelta(td, '{D:2}d {H:2}:{M:02}:{S:02}')
 2d  3:05:08

>>> print strfdelta(td, '{H}h {S}s')
51h 308s

>>> print strfdelta(12304, inputtype='s')
00d 03h 25m 04s

>>> print strfdelta(620, '{H}:{M:02}', 'm')

>>> print strfdelta(49, '{D}d {H}h', 'h')
2d 1h
Answered By: MarredCheese

If you already have a timedelta obj then just convert that obj into string. Remove the last 3 characters of the string and print. This will truncate the seconds part and print the rest of it in the format Hours:Minutes.

t = str(timedeltaobj) 

print t[:-3]
Answered By: Sima
import datetime
hours = datetime.timedelta(hours=16, minutes=30)
print((datetime.datetime(1,1,1) + hours).strftime('%H:%M'))
Answered By: SleX

I used the humanfriendly python library to do this, it works very well.

import humanfriendly
from datetime import timedelta
delta = timedelta(seconds = 321)

'5 minutes and 21 seconds'

Available at

Answered By: Dhiraj Gupta

A straight forward template filter for this problem. The built-in function int() never rounds up. F-Strings (i.e. f”) require python 3.6.

def diffTime(end, start):
    diff = (end - start).total_seconds()
    d = int(diff / 86400)
    h = int((diff - (d * 86400)) / 3600)
    m = int((diff - (d * 86400 + h * 3600)) / 60)
    s = int((diff - (d * 86400 + h * 3600 + m *60)))
    if d > 0:
        fdiff = f'{d}d {h}h {m}m {s}s'
    elif h > 0:
        fdiff = f'{h}h {m}m {s}s'
    elif m > 0:
        fdiff = f'{m}m {s}s'
        fdiff = f'{s}s'
    return fdiff
Answered By: Patrick

One liner. Since timedeltas do not offer datetime’s strftime, bring the timedelta back to a datetime, and use stftime.

This can not only achieve the OP’s requested format Hours:Minutes, now you can leverage the full formatting power of datetime’s strftime, should your requirements change to another representation.

import datetime
td = datetime.timedelta(hours=2, minutes=10, seconds=5)
print(datetime.datetime.strftime(datetime.datetime.strptime(str(td), "%H:%M:%S"), "%H:%M"))


This also solves the annoyance that timedeltas are formatted into strings as H:MM:SS rather than HH:MM:SS, which lead me to this problem, and the solution I’ve shared.

Answered By: Bill Gale

If you happen to have IPython in your packages (you should), it has (up to now, anyway) a very nice formatter for durations (in float seconds). That is used in various places, for example by the %%time cell magic. I like the format it produces for short durations:

>>> from IPython.core.magics.execution import _format_time
>>> for v in range(-9, 10, 2):
...     dt = 1.25 * 10**v
...     print(_format_time(dt))

1.25 ns
125 ns
12.5 µs
1.25 ms
125 ms
12.5 s
20min 50s
1d 10h 43min 20s
144d 16h 13min 20s
14467d 14h 13min 20s
Answered By: Pierre D

Here’s a function to stringify timedelta.total_seconds(). It works in python 2 and 3.

def strf_interval(seconds):
    days, remainder = divmod(seconds, 86400)
    hours, remainder = divmod(remainder, 3600)
    minutes, seconds = divmod(remainder, 60)
    return '{} {} {} {}'.format(
            "" if int(days) == 0 else str(int(days)) + ' days',
            "" if int(hours) == 0 else str(int(hours)) + ' hours',
            "" if int(minutes) == 0 else str(int(minutes))  + ' mins',
            "" if int(seconds) == 0 else str(int(seconds))  + ' secs'

Example output:

>>> print(strf_interval(1))
   1 secs
>>> print(strf_interval(100))
  1 mins 40 secs
>>> print(strf_interval(1000))
  16 mins 40 secs
>>> print(strf_interval(10000))
 2 hours 46 mins 40 secs
>>> print(strf_interval(100000))
1 days 3 hours 46 mins 40 secs
Answered By: Yuriy Mankovskiy

I have a function:

def period(delta, pattern):
    d = {'d': delta.days}
    d['h'], rem = divmod(delta.seconds, 3600)
    d['m'], d['s'] = divmod(rem, 60)
    return pattern.format(**d)


>>> td = timedelta(seconds=123456789)
>>> period(td, "{d} days {h}:{m}:{s}")
'1428 days 21:33:9'
>>> period(td, "{h} hours, {m} minutes and {s} seconds, {d} days")
'21 hours, 33 minutes and 9 seconds, 1428 days'
Answered By: GooDeeJAY

I continued from MarredCheese’s answer and added year, month, millicesond and microsecond

all numbers are formatted to integer except for second, thus the fraction of a second can be customized.

@kfmfe04 asked for fraction of a second so I posted this solution

In the main there are some examples.

from string import Formatter
from datetime import timedelta

def strfdelta(tdelta, fmt='{D:02}d {H:02}h {M:02}m {S:02.0f}s', inputtype='timedelta'):
    """Convert a datetime.timedelta object or a regular number to a custom-
    formatted string, just like the stftime() method does for datetime.datetime

    The fmt argument allows custom formatting to be specified.  Fields can 
    include seconds, minutes, hours, days, and weeks.  Each field is optional.

    Some examples:
        '{D:02}d {H:02}h {M:02}m {S:02.0f}s' --> '05d 08h 04m 02s' (default)
        '{W}w {D}d {H}:{M:02}:{S:02.0f}'     --> '4w 5d 8:04:02'
        '{D:2}d {H:2}:{M:02}:{S:02.0f}'      --> ' 5d  8:04:02'
        '{H}h {S:.0f}s'                       --> '72h 800s'

    The inputtype argument allows tdelta to be a regular number instead of the  
    default, which is a datetime.timedelta object.  Valid inputtype strings: 
        's', 'seconds', 
        'm', 'minutes', 
        'h', 'hours', 
        'd', 'days', 
        'w', 'weeks'

    # Convert tdelta to integer seconds.
    if inputtype == 'timedelta':
        remainder = tdelta.total_seconds()
    elif inputtype in ['s', 'seconds']:
        remainder = float(tdelta)
    elif inputtype in ['m', 'minutes']:
        remainder = float(tdelta)*60
    elif inputtype in ['h', 'hours']:
        remainder = float(tdelta)*3600
    elif inputtype in ['d', 'days']:
        remainder = float(tdelta)*86400
    elif inputtype in ['w', 'weeks']:
        remainder = float(tdelta)*604800

    f = Formatter()
    desired_fields = [field_tuple[1] for field_tuple in f.parse(fmt)]
    possible_fields = ('Y','m','W', 'D', 'H', 'M', 'S', 'mS', 'µS')
    constants = {'Y':86400*365.24,'m': 86400*30.44 ,'W': 604800, 'D': 86400, 'H': 3600, 'M': 60, 'S': 1, 'mS': 1/pow(10,3) , 'µS':1/pow(10,6)}
    values = {}
    for field in possible_fields:
        if field in desired_fields and field in constants:
            Quotient, remainder = divmod(remainder, constants[field])
            values[field] = int(Quotient) if field != 'S' else Quotient + remainder
    return f.format(fmt, **values)

if __name__ == "__main__":
    td = timedelta(days=717, hours=3, minutes=5, seconds=8, microseconds=3549)
    print(strfdelta(td,'{Y} years {m} months {W} weeks {D} days {H:02}:{M:02}:{S:02}'))  
    print(strfdelta(td,'{m} months {W} weeks {D} days {H:02}:{M:02}:{S:02.4f}'))  
    td = timedelta( seconds=8, microseconds=8549)
    print(strfdelta(td,'{S} seconds {mS} milliseconds {µS} microseconds'))  
    print(strfdelta(td,'{S:.0f} seconds {mS} milliseconds {µS} microseconds'))  


1 years 11 months 2 weeks 3 days 01:09:56.00354900211096
23 months 2 weeks 3 days 00:12:20.0035
8.008549 seconds 8 milliseconds 549 microseconds
8 seconds 8 milliseconds 549 microseconds
115d 17h 46m 40s
Answered By: tomatoeshift

timedelta to string, use for print running time info.

def strfdelta_round(tdelta, round_period='second'):
  """timedelta to string,  use for measure running time
  attend period from days downto smaller period, round to minimum period
  omit zero value period  
  period_names = ('day', 'hour', 'minute', 'second', 'millisecond')
  if round_period not in period_names:
    raise Exception(f'round_period "{round_period}" invalid, should be one of {",".join(period_names)}')
  period_seconds = (86400, 3600, 60, 1, 1/pow(10,3))
  period_desc = ('days', 'hours', 'mins', 'secs', 'msecs')
  round_i = period_names.index(round_period)
  s = ''
  remainder = tdelta.total_seconds()
  for i in range(len(period_names)):
    q, remainder = divmod(remainder, period_seconds[i])
    if int(q)>0:
      if not len(s)==0:
        s += ' '
      s += f'{q:.0f} {period_desc[i]}'
    if i==round_i:
    if i==round_i+1:
      s += f'{remainder} {period_desc[round_i]}'
  return s

e.g. auto omit zero leading period:

>>> td = timedelta(days=0, hours=2, minutes=5, seconds=8, microseconds=3549)
>>> strfdelta_round(td, 'second')
'2 hours 5 mins 8 secs'

or omit middle zero period:

>>> td = timedelta(days=2, hours=0, minutes=5, seconds=8, microseconds=3549)
>>> strfdelta_round(td, 'millisecond')
'2 days 5 mins 8 secs 3 msecs'

or round to minutes, omit below minutes:

>>> td = timedelta(days=1, hours=2, minutes=5, seconds=8, microseconds=3549)
>>> strfdelta_round(td, 'minute')
'1 days 2 hours 5 mins'
Answered By: raidsan


>>> import datetime
>>> dt0 = datetime.datetime(1,1,1)
>>> td = datetime.timedelta(minutes=34, hours=12, seconds=56)
>>> (dt0+td).strftime('%X')
>>> (dt0+td).strftime('%M:%S')
>>> (dt0+td).strftime('%H:%M')
Answered By: Mechanic

I suggest the following method so that we can utilize the standard formatting function, pandas.Timestamp.strftime!

from pandas import Timestamp, Timedelta

(Timedelta("2 hours 30 min") + Timestamp("00:00:00")).strftime("%H:%M")
Answered By: Sunghee Yun

I wanted to do this so wrote a simple function. It works great for me and is quite versatile (supports years to microseconds, and any granularity level, e.g. you can pick between ‘2 days, 4 hours, 48 minutes’ and ‘2 days, 4 hours’ and ‘2 days, 4.8 hours’, etc.

def pretty_print_timedelta(t, max_components=None, max_decimal_places=2):
Print a pretty string for a timedelta. 
For example datetime.timedelta(days=2, seconds=17280) will be printed as '2 days, 4 hours, 48 minutes'. Setting max_components to e.g. 1 will change this to '2.2 days', where the 
number of decimal points can also be set. 
time_scales = [timedelta(days=365), timedelta(days=1), timedelta(hours=1), timedelta(minutes=1), timedelta(seconds=1), timedelta(microseconds=1000), timedelta(microseconds=1)]
time_scale_names_dict = {timedelta(days=365): 'year',  
                         timedelta(days=1): 'day', 
                         timedelta(hours=1): 'hour', 
                         timedelta(minutes=1): 'minute', 
                         timedelta(seconds=1): 'second', 
                         timedelta(microseconds=1000): 'millisecond', 
                         timedelta(microseconds=1): 'microsecond'}
count = 0
txt = ''
first = True
for scale in time_scales:
    if t >= scale: 
        count += 1
        if count == max_components:
            n = t / scale
            n = int(t / scale)
        t -= n*scale
        n_txt = str(round(n, max_decimal_places))
        if n_txt[-2:]=='.0': n_txt = n_txt[:-2]
        txt += '{}{} {}{}'.format('' if first else ', ', n_txt, time_scale_names_dict[scale], 's' if n>1 else '', )
        if first:
            first = False
if len(txt) == 0: 
    txt = 'none'
return txt
Answered By: Edden Gerber
# Format seconds to days, hours, minutes and seconds string
def ptime(seconds):
if(seconds >= 86400):
    d = seconds // 86400 # // floor division
    return (f"{round(d)}d") + ptime(seconds - d * 86400)
    if(seconds >= 3600):
        h = seconds // 3600 
        return (f"{round(h)}h") + ptime(seconds - h * 3600)
        if(seconds >= 60):
            m = seconds // 60
            return(f"{round(m)}m" + ptime(seconds - m * 60))
            if (seconds > 0):
Answered By: Manuel Alves

I had the same problem and I am using pandas Timedeltas, didn’t want to bring in additional dependencies (another answer mentions humanfriendly) so I wrote this small function to print out only the relevant information:

def format_timedelta(td: pd.Timedelta) -> str:
    if pd.isnull(td):
        return str(td)
        c = td.components._asdict()
        return ", ".join(f"{n} {unit}" for unit, n in c.items() if n)

For example, pd.Timedelta(hours=3, seconds=12) would print as 3 hours, 12 seconds.

Answered By: ilCatania