如果他们有共同的段，加入行列表？

Question

我有行列表，每一行都是这样的列表

[firstpoint, secondpoint]

例如我有 4 行是这样的：

listoflines = [[0,2],[1,3],[2,5],[6,7]]

所以当你看到一些线有相互段时，我想要的是如果它们有相互段就加入它们，因此结果会是这样的：

newlistoflines = [[0,5],[6,7]]

我尝试的是使用一个函数将它们相互比较并循环遍历我的行列表，但我对其结果有疑问：

def JOINLINES(line1, line2):
    x1 = line1[0]
    x2 = line1[1]
    x3 = line2[0]
    x4 = line2[1]

    if x2 < x3:
        result = (x1, x2), (x3, x4)
    else:
        result = (min(x1, x2, x3, x4), max(x1, x2, x3, x4))

    return result

newbeamlist = []
for i in range(1, len(beams)):
    newbeamlist.append(JOINLINES(beams[i - 1], beams[i]))

输出=[(0, 3), (1, 5), ((2, 5), (6, 7))]

Answer 1

您的解决方案是附加 JOINLINES 函数的结果。

不考虑下一条也可以是互段

试试这个。我假设了一些事情。

# assuming that the items in the list are ordered according to the first item in each sublist. 
# This will enable us to find mutual segments (overlaps) easily
# if the list is not sorted, then we can sort it using the first item in each sublist item

li = [[0,2],[1,3],[2,5],[6,7], [10,11],[7,9]]

# sorting using the first item in each sublist item
li.sort(key=lambda x: x[0])
# out: [[0,5],[6,7]]
newli= []
left_prev = li[0][0]
right_prev = li[0][1]
for i in range(1,len(li)):
    left_current = li[i][0]
    right_current = li[i][1]
    if left_current <= right_prev:
        # mutual segment (overlap) found
        right_prev = right_current
        continue
    else:
        # no mutual segment(overlap), append the previous to the new list
        newli.append([left_prev, right_prev])
        left_prev = left_current
        right_prev = right_current

# appending the last item
newli.append([left_prev, right_prev])
print(newli) # [[0, 5], [6, 9], [10, 11]]

如果他们有共同的段，加入行列表？

joining list of lines if they have mutual segments?

python

list

join