Question

我将持续时间表示为一对像这样的日期时间对象

(datetime(2014, 1, 23, 14, 49, 6, 185955), datetime(2015, 4, 18, 14, 49, 6, 185955))

我有一个像这样的元组列表

[(datetime(2014, 1, 23, 14, 49, 6, 185955), datetime(2015, 4, 18, 14, 49, 6, 185955)), (datetime(2015, 2, 2, 14, 49, 6, 185955), datetime(2015, 10, 23, 14, 49, 6, 185955)),]

现在如何计算列表中所有持续时间（此处为datetime对象对）的（小时，分钟，秒）重叠持续时间？

E.g：

A        C                B         D    G   E          H   I     J  F
|-----------------------|
         |--------------------------|
                                             |-----------------------|
                                         |--------------|   |-----|

这里AB，CD是一对日期时间对象。如何在（小时，分钟，秒）中获得重叠CB 这样的一对列表？

Answer 1

试试这个：

from datetime import datetime
t = [(datetime(2014, 1, 23, 14, 49, 6, 185955), datetime(2015, 4, 18, 14, 49, 6, 185955)), (datetime(2015, 2, 2, 14, 49, 6, 185955), datetime(2015, 10, 23, 14, 49, 6, 185955))]
def time_overlap(t):
    overlap = (min(t[0][1], t[1][1]) - max(t[0][0], t[1][0]))
    days = overlap.days+1
    hours, remainder = divmod(overlap.seconds, 3600)
    minutes, seconds = divmod(remainder, 60)
    hours += days*24
    return (hours,minutes,seconds)
print(time_overlap(t))

输出：

(1824, 0, 0)

要查找多个日期时间的重叠，请执行以下操作：

from datetime import datetime
t = [(datetime(2014, 1, 23, 14, 49, 6, 185955), datetime(2015, 4, 18, 14, 49, 6, 185955)), (datetime(2015, 2, 2, 14, 49, 6, 185955), datetime(2015, 10, 23, 14, 49, 6, 185955))]
def time_overlap(t):
    overlap = (min([x[1] for x in t]) - max([x[0] for x in t]))
    days = overlap.days+1
    hours, remainder = divmod(overlap.seconds, 3600)
    minutes, seconds = divmod(remainder, 60)
    hours += days*24
    return (hours,minutes,seconds)
print(time_overlap(t))

Python - 如何找到日期时间重叠？

1 个答案: