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'

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)

我接着MarredCheese的回答，加上了年、月、毫秒和微秒

除秒外，所有数字都被格式化为整数，因此秒的分数可以自定义。

@kfmfe04要求几分之一秒，所以我发布了这个解决方案

大体上有一些例子。

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
    objects.

    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'))  
    print(strfdelta(pow(10,7),inputtype='s'))

输出:

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

>>> str(datetime.timedelta(hours=10.56))
10:33:36

>>> td = datetime.timedelta(hours=10.505) # any timedelta object
>>> ':'.join(str(td).split(':')[:2])
10:30

将timedelta对象传递给str()函数调用的格式化代码与输入print td相同。因为你不想要秒，我们可以用冒号分隔字符串(3部分)，然后只用前2部分把它重新组合在一起。

我建议使用以下方法，这样我们就可以利用标准格式化函数pandas.Timestamp.strftime!

from pandas import Timestamp, Timedelta

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