按照上面Joe的示例值，我将使用模算术运算符，因此:

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

注意，Python中的整数除法默认是四舍五入;如果想要更显式，可以适当使用math.floor()或math.ceil()。

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)

这将产生:

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

如果你已经有一个timedelta obj，那么只需将obj转换为字符串。删除字符串的最后3个字符并打印。这将截断秒部分，并以小时:分钟的格式打印其余部分。

t = str(timedeltaobj) 

print t[:-3]

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

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")