下面是一个stringify timedelta.total_seconds()的函数。它在python2和python3中工作。

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'
        )

示例输出:

>>> 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

我知道这是一个老问题，但我使用datetime.utcfromtimestamp()来解决这个问题。它接受秒数并返回一个datetime，该datetime可以像任何其他datetime一样格式化。

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

只要您停留在时间部分的合法范围内，这就应该工作，即它不会返回1234:35，因为小时<= 23。

如您所知，您可以通过访问.seconds属性从timedelta对象中获得total_seconds。

Python提供了内置函数divmod()，它允许:

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

或者你可以结合使用模和减法来转换小时和余数:

# 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

我想这样做，所以写了一个简单的函数。它对我来说非常有用，而且非常通用(支持年到微秒，以及任何粒度级别，例如，你可以选择“2天4小时48分钟”和“2天4小时”和“2天4.8小时”等。

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
        else:
            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

我使用人类友好的python库来做这件事，它工作得非常好。

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

'5 minutes and 21 seconds'

网址:https://pypi.org/project/humanfriendly/

谢谢大家的帮助。我把你的很多想法放在一起，让我知道你的想法。

我像这样在类中添加了两个方法:

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

在我的django中，我使用了这个(sum是对象，它在一个字典中):

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