试试这个方法吧——它对安排短期事件很有效。如果某些内容花费了一个多小时，那么最终的显示可能需要一些友好的格式。

import time
start = time.time()

time.sleep(10)  # or do something more productive

done = time.time()
elapsed = done - start
print(elapsed)

时间差以经过的秒数返回。

这个网站建议你试试:

import datetime as dt
start="09:35:23"
end="10:23:00"
start_dt = dt.datetime.strptime(start, '%H:%M:%S')
end_dt = dt.datetime.strptime(end, '%H:%M:%S')
diff = (end_dt - start_dt) 
diff.seconds/60 

这个论坛使用time.mktime()

看一下datetime模块和timedelta对象。您最终应该为开始和停止时间构造一个datetime对象，当您减去它们时，您将得到一个timedelta。

    import datetime
    
    day = int(input("day[1,2,3,..31]: "))
    month = int(input("Month[1,2,3,...12]: "))
    year = int(input("year[0~2020]: "))
    start_date = datetime.date(year, month, day)
    
    day = int(input("day[1,2,3,..31]: "))
    month = int(input("Month[1,2,3,...12]: "))
    year = int(input("year[0~2020]: "))
    end_date = datetime.date(year, month, day)
    
    time_difference = end_date - start_date
    age = time_difference.days
    print("Total days: " + str(age))

如果你只对24小时以内的时间感兴趣，那就简明扼要。你可以在return语句中根据需要格式化输出:

import datetime
def elapsed_interval(start,end):
    elapsed = end - start
    min,secs=divmod(elapsed.days * 86400 + elapsed.seconds, 60)
    hour, minutes = divmod(min, 60)
    return '%.2d:%.2d:%.2d' % (hour,minutes,secs)

if __name__ == '__main__':
    time_start=datetime.datetime.now()
    """ do your process """
    time_end=datetime.datetime.now()
    total_time=elapsed_interval(time_start,time_end)

下面是如果字符串也包含天数的代码[-1 day 32:43:02]:

print(
    (int(time.replace('-', '').split(' ')[0]) * 24) * 60 
    + (int(time.split(' ')[-1].split(':')[0]) * 60)
    + int(time.split(' ')[-1].split(':')[1])
)