python datetime.time 操作

Question

t1 = datetime.time(12, 10, 0, tzinfo=GMT1()) # 12:10
t2 = datetime.time(13, 13, 0, tzinfo=GMT1()) #13:13
t3 = datetime.time(23, 55, 0, tzinfo=GMT1()) #23:55 
t4 =  datetime.time(01, 10, 0, tzinfo=GMT1()) #01:10

我需要两次之间的分钟间隔。例如，一个非工作时间：

def minute_interval(start,end):
     return end - start

minute_interval(t1,t2) #should give 63 mins.

此外，如果结束时间小于开始时间，则应该通过了解结束时间是从第二天的时间开始来进行计算。即：

minute_interval(t3,t4) #should give 75 mins.

如何实现这一点？为此，我需要重写 分钟间隔 函数。

Answer 1

假设时间位于同一时区并且没有 DST

import datetime

def minute_interval(start, end):
     reverse = False
     if start > end:
          start, end = end, start
          reverse = True

     delta = (end.hour - start.hour)*60 + end.minute - start.minute + (end.second - start.second)/60.0
     if reverse:
          delta = 24*60 - delta
     return delta

t1 = datetime.time(12, 10, 0) # 12:10
t2 = datetime.time(13, 13, 0) #13:13
t3 = datetime.time(23, 55, 0) #23:55 
t4 = end = datetime.time(01, 10, 0) #01:10

print minute_interval(t1, t2)
print minute_interval(t3, t4)

输出：

63
75

否则您最好使用 datetime.datetime，它支持减法并提供 datetime.timedelta，对于时区，您可以使用 pytz 库。

Answer 2

这怎么样？

def minute_interval(start,end):
    start_sec= (start.hour*60+start.minute)*60+start.second
    end_sec= (end.hour*60+end.minute)*60+end.second
    return (end_sec-start_sec)/60.0

Answer 3

>>> import datetime
>>> t1 = datetime.time(12, 10, 0)
>>> t2 = datetime.time(13, 13, 0)
>>> today = datetime.datetime.today()
>>> t1n = datetime.datetime.combine(today, t1)
>>> t2n = datetime.datetime.combine(today + datetime.timedelta(days=int(t2 < t1)), t2)
>>> d = t2n - t1n
>>> (d.days * 86400 + d.seconds) // 60
63