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

我有一个函数:

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)

例子:

>>> 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'

>>> 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")

提问者想要一个比典型的更好的格式:

  >>> import datetime
  >>> datetime.timedelta(seconds=41000)
  datetime.timedelta(0, 41000)
  >>> str(datetime.timedelta(seconds=41000))
  '11:23:20'
  >>> str(datetime.timedelta(seconds=4102.33))
  '1:08:22.330000'
  >>> str(datetime.timedelta(seconds=413302.33))
  '4 days, 18:48:22.330000'

所以，实际上有两种格式，一种是天数为0，它被省略了，另一种是文本“n天，h:m:s”。但是，秒可能有分数，打印输出中没有前导0，所以列很乱。

如果你喜欢，下面是我的日常工作:

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

返回dd:hh:mm:ss格式的输出:

00:00:00:15
00:00:00:19
02:01:31:40
02:01:32:22

我确实想过在上面加上年份，但这是留给读者的练习，因为超过1年的输出是安全的:

>>> str(datetime.timedelta(seconds=99999999))
'1157 days, 9:46:39'

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