用python连接两个列表列表

Question

我有两个列表

a = [[1,2],[5,3],[7,9]]
b = [[2,4], [6,7]]

我想将列表连接到

[[1,2,4],[5,3],[6,7,9]]

目标是如果列表中有相同的元素，它们将被连接起来。 非常感谢任何帮助。

Answer 1

这应该适用于更一般的情况：

def connected_components(list_of_lists):
    """ based on Howard's answer https://stackoverflow.com/a/4842897/10693596 """
    temp_list_copy = list_of_lists.copy()
    result = []
    while len(temp_list_copy)>0:
        first, *rest = temp_list_copy
        first = set(first)

        lf = -1
        while len(first)>lf:
            lf = len(first)

            rest2 = []
            for r in rest:
                if len(first.intersection(set(r)))>0:
                    first |= set(r)
                else:
                    rest2.append(r)     
            rest = rest2
        result.append(list(first))
        temp_list_copy = rest
    return result


a = [[1,2],[5,3],[7,9]]
b = [[2,4], [6,7]]

a = connected_components(a)
b = connected_components(b)

for n, i in enumerate(a):
    combined_list = a[n]+ [jj for j in b if set(j).intersection(set(i)) for jj in j]

    a[n] = sorted(list(set(combined_list)))

print(a)

或者也许下面是一个更Pythonic的版本：

result = [
    sorted(
        set([
            k
            for j in b
            for k in (set(i)|set(j) if set(i)&set(j) else set(i))
        ])
    )
    for i in a
]
print(result)

Answer 2

这是一个使用集合的解决方案： 它适用于 OP 示例以及评论中给出的示例。

a = [[1,2], [2,3],[7,9]]
b = [[2,4], [6,7]]     

a = a+b
b = []

while a != []:
    i = a.pop()
    for j in range(len(b)):
        if set(b[j]).intersection(set(i)) != set():
            b[j] = list(set(b[j]).union(set(i)))
            break
    else:
        if i != []:
            b.append(i)


print(b)
## [[9, 6, 7], [1, 2, 3, 4]]

其他测试：

a = [[8, 9], [1,2],[5,3],[7,9], [5, 6]]
b = [[2,4], [6,7]]
## [[3, 5, 6, 7, 8, 9], [1, 2, 4]]

a = [[1,2], [2,3],[7,9]]
b = [[2,4], [6,7]]
## [[9, 6, 7], [1, 2, 3, 4]]