可能:

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

如果您的包中碰巧有IPython(您应该)，那么它(到目前为止)有一个非常好的持续时间格式器(以浮点秒为单位)。它被用于许多地方，例如%%时间单元格魔法。我喜欢它在短时间内产生的格式:

>>> 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

您可以使用str()将timedelta转换为字符串。这里有一个例子:

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

我在工作中遇到过类似的加班计算输出问题。该值应该始终以HH:MM显示，即使它大于一天并且该值可能为负值。我结合了一些展示的解决方案，也许其他人会发现这个解决方案很有用。我意识到，如果timedelta值为负，大多数divmod方法所显示的解决方案都不能开箱即用:

def td2HHMMstr(td):
  '''Convert timedelta objects to a HH:MM string with (+/-) sign'''
  if td < datetime.timedelta(seconds=0):
    sign='-'
    td = -td
  else:
    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

td2HHMMstr(datetime.timedelta(hours=1, minutes=45))
'1:54'

td2HHMMstr(datetime.timedelta(days=2, hours=3, minutes=2))
'51:02'

td2HHMMstr(datetime.timedelta(hours=-3, minutes=-2))
'-3:02'

td2HHMMstr(datetime.timedelta(days=-35, hours=-3, minutes=-2))
'-843:02'

针对这个问题的一个直接的模板过滤器。内置函数int()从不四舍五入。f - string(即f'')需要python 3.6。

@app_template_filter()
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'
    else:
        fdiff = f'{s}s'
    return fdiff

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)
    ]

    strings=[]
    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)