import datetime
date = datetime.date(1, 1, 1)
#combine a dummy date to the time
datetime1 = datetime.datetime.combine(date, start_time)
datetime2 = datetime.datetime.combine(date, stop_time)  
#compute the difference
time_elapsed = datetime1 - datetime2

Start_time——> datetime对象的开始时间 End_time——> datetime对象的结束时间

我们不能直接减去datetime。时间对象 因此，我们需要添加一个随机日期(我们使用combine) 或者你可以用“today”来代替(1,1,1)

希望这能有所帮助

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”

我利用时差进行了连续集成测试，以检查和完善我的功能。如果有人需要，这里有一些简单的代码

from datetime import datetime

class TimeLogger:
    time_cursor = None

    def pin_time(self):
        global time_cursor
        time_cursor = datetime.now()

    def log(self, text=None) -> float:
        global time_cursor

        if not time_cursor:
            time_cursor = datetime.now()

        now = datetime.now()
        t_delta = now - time_cursor

        seconds = t_delta.total_seconds()

        result = str(now) + ' tl -----------> %.5f' % seconds
        if text:
            result += "   " + text
        print(result)

        self.pin_time()

        return seconds


time_logger = TimeLogger()

使用:

from .tests_time_logger import time_logger
class Tests(TestCase):
    def test_workflow(self):
    time_logger.pin_time()

    ... my functions here ...

    time_logger.log()

    ... other function(s) ...

    time_logger.log(text='Tests finished')

我在log输出中有类似的东西

2019-12-20 17:19:23.635297 tl -----------> 0.00007
2019-12-20 17:19:28.147656 tl -----------> 4.51234   Tests finished

只要用一个减去另一个。你会得到一个timedelta对象。

>>> import datetime
>>> d1 = datetime.datetime.now()
>>> d2 = datetime.datetime.now() # after a 5-second or so pause
>>> d2 - d1
datetime.timedelta(0, 5, 203000)
>>> dd = d2 - d1
>>> print (dd.days) # get days
>>> print (dd.seconds) # get seconds
>>> print (dd.microseconds) # get microseconds
>>> print (int(round(dd.total_seconds()/60, 0))) # get minutes

基于@Attaque的精彩回答，我提出了一个更短的简化版本的日期时差计算器:

seconds_mapping = {
    'y': 31536000,
    'm': 2628002.88, # this is approximate, 365 / 12; use with caution
    'w': 604800,
    'd': 86400,
    'h': 3600,
    'min': 60,
    's': 1,
    'mil': 0.001,
}

def get_duration(d1, d2, interval, with_reminder=False):
    if with_reminder:
        return divmod((d2 - d1).total_seconds(), seconds_mapping[interval])
    else:
        return (d2 - d1).total_seconds() / seconds_mapping[interval]

我改变了它，以避免声明重复的函数，删除了漂亮的打印默认间隔，并增加了对毫秒、周和ISO月的支持(简单地说，月只是一个近似，基于每个月等于365/12的假设)。

生产:

d1 = datetime(2011, 3, 1, 1, 1, 1, 1000)
d2 = datetime(2011, 4, 1, 1, 1, 1, 2500)

print(get_duration(d1, d2, 'y', True))      # => (0.0, 2678400.0015)
print(get_duration(d1, d2, 'm', True))      # => (1.0, 50397.12149999989)
print(get_duration(d1, d2, 'w', True))      # => (4.0, 259200.00149999978)
print(get_duration(d1, d2, 'd', True))      # => (31.0, 0.0014999997802078724)
print(get_duration(d1, d2, 'h', True))      # => (744.0, 0.0014999997802078724)
print(get_duration(d1, d2, 'min', True))    # => (44640.0, 0.0014999997802078724)
print(get_duration(d1, d2, 's', True))      # => (2678400.0, 0.0014999997802078724)
print(get_duration(d1, d2, 'mil', True))    # => (2678400001.0, 0.0004999997244524721)

print(get_duration(d1, d2, 'y', False))     # => 0.08493150689687975
print(get_duration(d1, d2, 'm', False))     # => 1.019176965856293
print(get_duration(d1, d2, 'w', False))     # => 4.428571431051587
print(get_duration(d1, d2, 'd', False))     # => 31.00000001736111
print(get_duration(d1, d2, 'h', False))     # => 744.0000004166666
print(get_duration(d1, d2, 'min', False))   # => 44640.000024999994
print(get_duration(d1, d2, 's', False))     # => 2678400.0015
print(get_duration(d1, d2, 'mil', False))   # => 2678400001.4999995

要查找天数:timedelta有一个'days'属性。你可以简单地查询一下。

>>>from datetime import datetime, timedelta
>>>d1 = datetime(2015, 9, 12, 13, 9, 45)
>>>d2 = datetime(2015, 8, 29, 21, 10, 12)
>>>d3 = d1- d2
>>>print d3
13 days, 15:59:33
>>>print d3.days
13