>>> import datetime
>>> first_time = datetime.datetime.now()
>>> later_time = datetime.datetime.now()
>>> difference = later_time - first_time
datetime.timedelta(0, 8, 562000)
>>> seconds_in_day = 24 * 60 * 60
>>> divmod(difference.days * seconds_in_day + difference.seconds, 60)
(0, 8)      # 0 minutes, 8 seconds

从第一个时间差中减去后面的时间= later_time - first_time创建一个只保存时间差的datetime对象。 在上面的例子中，它是0分钟，8秒和562000微秒。

只是认为在timedelta方面提到格式化可能是有用的。Strptime()根据格式解析表示时间的字符串。

from datetime import datetime

datetimeFormat = '%Y/%m/%d %H:%M:%S.%f'    
time1 = '2016/03/16 10:01:28.585'
time2 = '2016/03/16 09:56:28.067'  
time_dif = datetime.strptime(time1, datetimeFormat) - datetime.strptime(time2,datetimeFormat)
print(time_dif)

这将输出: 0:05:00.518000

要得到小时、分和秒，你可以这样做

>>> import datetime
>>> first_time = datetime.datetime.now()
>>> later_time = datetime.datetime.now()
>>> difference = later_time - first_time
>>> m, s = divmod(difference.total_seconds(), 60)
>>> print("H:M:S is {}:{}:{}".format(m//60, m%60, s))

这里有一个很容易概括或转化为函数的答案，它合理紧凑，易于遵循。

ts_start=datetime(2020, 12, 1, 3, 9, 45)
ts_end=datetime.now()
ts_diff=ts_end-ts_start
secs=ts_diff.total_seconds()
days,secs=divmod(secs,secs_per_day:=60*60*24)
hrs,secs=divmod(secs,secs_per_hr:=60*60)
mins,secs=divmod(secs,secs_per_min:=60)
secs=round(secs, 2)
answer='Duration={} days, {} hrs, {} mins and {} secs'.format(int(days),int(hrs),int(mins),secs)
print(answer)

它给出的答案是“持续时间=270天10小时32分42.13秒”

这就是我如何获得两个datetime之间经过的小时数。datetime对象:

before = datetime.datetime.now()
after  = datetime.datetime.now()
hours  = math.floor(((after - before).seconds) / 3600)

我用的是这样的:

from datetime import datetime

def check_time_difference(t1: datetime, t2: datetime):
    t1_date = datetime(
        t1.year,
        t1.month,
        t1.day,
        t1.hour,
        t1.minute,
        t1.second)

    t2_date = datetime(
        t2.year,
        t2.month,
        t2.day,
        t2.hour,
        t2.minute,
        t2.second)

    t_elapsed = t1_date - t2_date

    return t_elapsed

# usage 
f = "%Y-%m-%d %H:%M:%S+01:00"
t1 = datetime.strptime("2018-03-07 22:56:57+01:00", f)
t2 = datetime.strptime("2018-03-07 22:48:05+01:00", f)
elapsed_time = check_time_difference(t1, t2)

print(elapsed_time)
#return : 0:08:52