在这里，我会认真考虑奥卡姆剃刀方法:

td = str(timedelta).split('.')[0]

这将返回一个没有微秒的字符串

如果要重新生成datetime。Timedelta对象，只需要这样做:

h,m,s = re.split(':', td)
new_delta = datetime.timedelta(hours=int(h),minutes=int(m),seconds=int(s))

2年了，我爱上了这门语言!

请检查这个函数-它将timedelta对象转换为字符串'HH:MM:SS'

def format_timedelta(td):
    hours, remainder = divmod(td.total_seconds(), 3600)
    minutes, seconds = divmod(remainder, 60)
    hours, minutes, seconds = int(hours), int(minutes), int(seconds)
    if hours < 10:
        hours = '0%s' % int(hours)
    if minutes < 10:
        minutes = '0%s' % minutes
    if seconds < 10:
        seconds = '0%s' % seconds
    return '%s:%s:%s' % (hours, minutes, seconds)

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

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'

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。

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