我有两个datetime.time
值,exit
并且enter
我想做类似的事情:
duration = exit - enter
但是,我收到此错误:
TypeError: 不支持的操作数类型 -: 'datetime.time' 和 'datetime.time
我该如何正确地做到这一点?一种可能的解决方案是将time
变量转换为datetime
变量,然后进行减法运算,但我相信你们一定有更好更干净的方法。
尝试这个:
from datetime import datetime, date
datetime.combine(date.today(), exit) - datetime.combine(date.today(), enter)
combine
构建一个可以减去的日期时间。
利用:
from datetime import datetime, date
duration = datetime.combine(date.min, end) - datetime.combine(date.min, beginning)
使用date.min
更加简洁,即使在午夜也可以使用。
date.today()
如果第一次调用发生在 23:59:59 并且下一次调用发生在 00:00:00,则可能会返回意外结果。
而不是使用时间尝试 timedelta:
from datetime import timedelta
t1 = timedelta(hours=7, minutes=36)
t2 = timedelta(hours=11, minutes=32)
t3 = timedelta(hours=13, minutes=7)
t4 = timedelta(hours=21, minutes=0)
arrival = t2 - t1
lunch = (t3 - t2 - timedelta(hours=1))
departure = t4 - t3
print(arrival, lunch, departure)
The python timedelta library should do what you need. A timedelta
is returned when you subtract two datetime
instances.
import datetime
dt_started = datetime.datetime.utcnow()
# do some stuff
dt_ended = datetime.datetime.utcnow()
print((dt_ended - dt_started).total_seconds())
您有两个 datetime.time 对象,因此您只需使用 datetime.timedetla 创建两个 timedelta,然后像现在一样使用“-”操作数进行减法。以下是不使用日期时间进行两次减法的示例方法。
enter = datetime.time(hour=1) # Example enter time
exit = datetime.time(hour=2) # Example start time
enter_delta = datetime.timedelta(hours=enter.hour, minutes=enter.minute, seconds=enter.second)
exit_delta = datetime.timedelta(hours=exit.hour, minutes=exit.minute, seconds=exit.second)
difference_delta = exit_delta - enter_delta
difference_delta是您的差异,您可以根据自己的原因使用它。
datetime.time
做不到 - 但你可以使用datetime.datetime.now()
start = datetime.datetime.now()
sleep(10)
end = datetime.datetime.now()
duration = end - start
import datetime
def diff_times_in_seconds(t1, t2):
# caveat emptor - assumes t1 & t2 are python times, on the same day and t2 is after t1
h1, m1, s1 = t1.hour, t1.minute, t1.second
h2, m2, s2 = t2.hour, t2.minute, t2.second
t1_secs = s1 + 60 * (m1 + 60*h1)
t2_secs = s2 + 60 * (m2 + 60*h2)
return( t2_secs - t1_secs)
# using it
diff_times_in_seconds( datetime.datetime.strptime( "13:23:34", '%H:%M:%S').time(),datetime.datetime.strptime( "14:02:39", '%H:%M:%S').time())
datetime.time
不支持这一点,因为以这种方式减去时间几乎没有意义。datetime.datetime
如果您想这样做,请使用完整的。
我和你有类似的情况,我最终使用了名为arrow的外部库。
这是它的样子:
>>> import arrow
>>> enter = arrow.get('12:30:45', 'HH:mm:ss')
>>> exit = arrow.now()
>>> duration = exit - enter
>>> duration
datetime.timedelta(736225, 14377, 757451)
timedelta
接受负(-)时间值。所以它可以简单如下(单行)
duration = datetime.timedelta(hours=exit.hour-enter.hour, minutes=exit.minute-enter.minute)
运行测试
import datetime
enter = datetime.time(hour=1, minute=30)
exit = datetime.time(hour=2, minute=0)
duration = datetime.timedelta(hours=exit.hour-enter.hour, minutes=exit.minute-enter.minute)
>>> duration
datetime.timedelta(seconds=1800)