Question

def mergeoverlapping(initialranges):
    i = sorted(set([tuple(sorted(x)) for x in initialranges]))

    f = [i[0]]
    for c, d in i[1:]:
        a, b = f[-1]
        if c<=b<d:
            f[-1] = a, d
        elif b<c<d:
            f.append((c,d))
        else:
            pass
    return f

def main():
    #open file for reading
    list_of_Tups = []
    with open("intervals.txt") as in_file:
        for line in in_file:
            int_list = [int(i) for i in line.split()]
            line = int_list
            list_of_Tups.append(line)
            list_of_Tups.sort()

    answer = list(mergeoverlapping(list_of_Tups))
    print("Non-intersecting Intervals:")
    for i in range (len(answer)):
        print(answer[i])

main()

给定一个数据文件，我必须创建元组，在列表中存储元组，对列表进行排序，并用一个元组替换每对重叠的元组。然后打印非交叉元组的列表。

但现在我想知道如何按照间隔大小的递增顺序打印非交叉间隔。如果两个间隔具有相同的大小，则按其下端的升序打印两个间隔。所以输出看起来像：

非交叉间隔：（-4,3）（4,7）（10,15）

按大小顺序的非交叉间隔：（4,7）（10,15）（-4,3）

Answer 1

假设：

intervals = [(-25, -14), (-10, -3), (2, 6), (12, 18), (22, 30)]

然后就像你已经排序一样，你只需根据大小进行排序。

# First sort in ascending order in relation to lower
intervals.sort(key=lambda interval: interval[0])

# Then sort in ascending order in relation to size
intervals.sort(key=lambda interval: abs(interval[0] - interval[1]))

然后print(intervals)输出：

[(2, 6), (12, 18), (-10, -3), (22, 30), (-25, -14)]

如果给定(-4, 3), (4, 7), (-2, 1)，则会产生(-2, 1), (4, 7), (-4, 3)（大小为3, 3, 7）。所以它正确地根据较低的尺寸进行排序。

如何以间隔大小的递增顺序打印非交叉间隔？

1 个答案: