Question

我有两个datetime范围列表。例如。

l1 = [(datetime.datetime(2018, 8, 29, 1, 0, 0), datetime.datetime(2018, 8, 29, 3, 0, 0)), (datetime.datetime(2018, 8, 29, 6, 0, 0), datetime.datetime(2018, 8, 29, 9, 0, 0))]
l2 = [(datetime.datetime(2018, 8, 29, 2, 0, 0), datetime.datetime(2018, 8, 29, 4, 0, 0)), (datetime.datetime(2018, 8, 29, 5, 0, 0), datetime.datetime(2018, 8, 29, 7, 0, 0))]

我想获得l1和l2的并集。所需的输出是：

union = [(datetime.datetime(2018, 8, 29, 1, 0, 0), datetime.datetime(2018, 8, 29, 4, 0, 0)), (datetime.datetime(2018, 8, 29, 5, 0, 0), datetime.datetime(2018, 8, 29, 9, 0, 0))]
intersection = [(datetime.datetime(2018, 8, 29, 2, 0, 0), datetime.datetime(2018, 8, 29, 3, 0, 0)), (datetime.datetime(2018, 8, 29, 6, 0, 0), datetime.datetime(2018, 8, 29, 7, 0, 0))]

实际数据可能无法完全对齐。

Answer 1

答案here对于您的要求非常有用，因为它可以压缩重叠范围的数组：

from operator import itemgetter

def consolidate(intervals):
    sorted_intervals = sorted(intervals, key=itemgetter(0))

    if not sorted_intervals:  # no intervals to merge
        return

    # low and high represent the bounds of the current run of merges
    low, high = sorted_intervals[0]

    for iv in sorted_intervals[1:]:
        if iv[0] <= high:  # new interval overlaps current run
            high = max(high, iv[1])  # merge with the current run
        else:  # current run is over
            yield low, high  # yield accumulated interval
            low, high = iv  # start new run

    yield low, high  # end the final run

l1和l2的结合只是l1和l2中所有范围的合并：

def union(l1, l2):
    return consolidate([*l1, *l2])

AChampion的代码足以完成l1和l2的交集（如果l1中的任何范围与l2中的任何范围之间都存在重叠，则该重叠值得结果），但可能导致范围碎片；我们可以使用相同的功能来加入相邻或重叠的范围：

from itertools import product

def intersection(l1, l2):
    result = ((max(s1, s2), min(e1, e2)) for (s1, e1), (s2, e2) in product(l1, l2) if s1 < e2 and e1 > s2)
    return consolidate(result)

一个例子：

l1 = [(1, 7), (4, 8), (10, 15), (20, 30), (50, 60)]
l2 = [(3, 6), (8, 11), (15, 20)]
print(list(union(l1, l2)))         # [(1, 30), (50, 60)]
print(list(intersection(l1, l2)))  # [(3, 6), (10, 11)]

（为清楚起见，该示例使用整数，但可与任何可比较的类型一起使用。具体地说，对于OP的l1和l2，代码将产生OP所需的datetime结果。）

Answer 2

您对日期范围的并集和交集的定义可以简单地描述为：-

联盟：

In []:
from itertools import product
[(min(s1, s2), max(e1, e2)) for (s1, e1), (s2, e2) in product(l1, l2) if s1 <= e2 and e1 >= s2]

Out[]:
[(datetime.datetime(2018, 8, 29, 1, 0), datetime.datetime(2018, 8, 29, 4, 0)),
 (datetime.datetime(2018, 8, 29, 5, 0), datetime.datetime(2018, 8, 29, 9, 0))]

交叉点：

In []:
[(max(s1, s2), min(e1, e2)) for (s1, e1), (s2, e2) in product(l1, l2) if s1 <= e2 and e1 >= s2]

Out[]:
[(datetime.datetime(2018, 8, 29, 2, 0), datetime.datetime(2018, 8, 29, 3, 0)),
 (datetime.datetime(2018, 8, 29, 6, 0), datetime.datetime(2018, 8, 29, 7, 0))]

如果<=和>=严格必须重叠而不仅仅是触摸，则可以用<和>替换{{1}}和{{1}}。

获取日期时间范围列表的并集和交集python

2 个答案: