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]

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)

t1 = datetime.datetime.strptime(StartTime, "%H:%M:%S %d-%m-%y")

t2 = datetime.datetime.strptime(EndTime, "%H:%M:%S %d-%m-%y")

return str(t2-t1)

所以对于:

StartTime = '15:28:53 21-07-13'
EndTime = '15:32:40 21-07-13'

返回:

'0:03:47'

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

他已经有一个timedelta对象，所以为什么不使用其内置方法total_seconds()将其转换为秒，然后使用divmod()获得小时和分钟?

hours, remainder = divmod(myTimeDelta.total_seconds(), 3600)
minutes, seconds = divmod(remainder, 60)

# Formatted only for hours and minutes as requested
print '%s:%s' % (hours, minutes)

无论时间增量是天还是年，这都是有效的。

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