我想输入一个区间列表并检查重叠区间并集的区间和非重叠区间的区间

Question

例如：我有 intervals = [[-5,-3],[-4,-1],[1,3],[4,8],[5,10],[10,12], [15,20]]（不一定像那样排序）

我想要return我[[-5,-1],[1,3],[4,12],[15,20]]的功能。因为 [-5,-3],[-4,-1] and [4,8],[5,10],[10,12] 有相互截取的数字。即，我希望函数 return 所有“孤独”间隔以及它们的数字相互截取的间隔并集的最小值和最大值。

我有这段代码可以做类似的事情，但这不是我想要的：

def maxDisjointIntervals(list_): 
      
    # Lambda function to sort the list   
    # elements by second element of pairs  
    list_.sort(key = lambda x: x[1]) 
      
    # First interval will always be  
    # included in set  
    print("[", list_[0][0], ", ", list_[0][1], "]") 
      
    # End point of first interval 
    r1 = list_[0][1] 
      
    for i in range(1, len(list_)): 
        l1 = list_[i][0] 
        r2 = list_[i][1] 
          
        # Check if given interval overlap with  
        # previously included interval, if not  
        # then include this interval and update  
        # the end point of last added interval 
        if l1 > r1: 
            print("[", l1, ", ", r2, "]") 
            r1 = r2

这段代码 return 给我这个输出：[[-5, -3],[1, 3],[4, 8],[10, 12],[15, 20]]

重复我之前说过的话：我希望它 return 我这个输出 [[-5, -1],[1, 3],[4, 12],[15, 20]]

我希望我的解释不会太冗长，因为我的母语不是英语。谢谢！

Answer 1

所以我的理解是你想找到 connected 或 overlapped 的间隔，你可以通过使用迭代器。

def maxDisjointIntervals(intervals):  # dont use list_ as your variable name
    overlappedIntervals = []

    if len(intervals) == 0:
        return overlappedIntervals

    # sort the intervals using the starting time
    intervals = sorted(intervals, key=lambda interval: interval[0])

    # init the start hour and end hour
    startHour = intervals[0][0]
    endHour = intervals[0][1]

    for interval in intervals[1:]:
        # if there is an overlap
        if interval[0] <= endHour <= interval[1]:
            endHour = interval[1]
        # if there is not an overlap
        else:
            overlappedIntervals.append([startHour, endHour])
            startHour = interval[0]
            endHour = interval[1]
    overlappedIntervals.append([startHour, endHour])
    return overlappedIntervals


print(maxDisjointIntervals([[-5, -3], [-4, -1], [1, 3], [4, 8], [5, 10], [10, 12], [13, 20]]))

输出：[[-5, -1], [1, 3], [4, 12], [13, 20]]

Answer 2

您可以使用 functools 中的 reduce 将区间合并在一起：

intervals = [[-5,-3],[-4,-1],[1,3],[4,8],[5,10],[10,12], [15,20]]

from functools import reduce
disjoints = [*reduce(lambda a,b: a+[b] if not a or b[0]>a[-1][1] else a[:-1]+[[a[-1][0],b[1]]],intervals,[])]

print(disjoints) # [[-5, -1], [1, 3], [4, 12], [15, 20]]

或者在基本循环中做同样的事情：

disjoints = intervals[:1]
for s,e in intervals[1:]:
    if s>disjoints[-1][-1]: disjoints.append([s,e])
    else: disjoints[-1][-1] = e
    
print(disjoints) # [[-5, -1], [1, 3], [4, 12], [15, 20]]

注意：这假定包含范围。如果末尾是独占使用 >= 而不是 >.

我想输入一个区间列表并检查重叠区间并集的区间和非重叠区间的区间

I want to input a list of intervals and check the intervals of the union of overlapping intervals and the intervals of non-overlapping intervals

python

set

intervals