如何将列表与列表列表中的公共元素合并？

Question

我正在尝试合并具有公共元素的列表列表中的所有列表。我有一些工作代码。但是，它似乎打破了这个例子：def merge_subs(lst_of_lsts):    res = []    for row in lst_of_lsts:        for i, resrow in enumerate(res):            if row[0]==resrow[0]:                res[i] += row[1:]                break            else:                res.append(sorted(row))    return sorted(res)输入是：merge_subs([[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]])我的结果是：[[0, 2, 4, 6], [1, 3, 5, 7], [3, 5, 7]]但我应该得到：[[0, 2, 4, 6], [1, 3, 5, 7]]

Answer 1

您需要使用递归：

def group(d, _start, _c = [], _seen = [], _used=[]):

  r = [i for i in d if any(c in _start for c in i) and i not in _seen and i not in _used]

  if not r:

    yield set(_c)

    for i in d:

      if i != _start and i not in _used:

         yield from group(d, i, _c=[], _seen=[], _used=_used+[i, *r])

  else:

    yield from group(d, _start, _c=_c+[i for b in r for i in b], _seen=_seen+r, _used=_used+r)

data = [[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]]

result = list(map(list, {tuple(i) for i in group(data, data[0], _seen=[data[0]]) if i}))

输出：

[[0, 2, 4, 6], [1, 3, 5, 7]]

Answer 2

这个问题可以使用递归解决，特别是尾递归：

def merge(lists, results=None):

    if results is None:

        results = []

    if not lists:

        return results

    first = lists[0]

    merged = []

    output = []

    for li in lists[1:]:

        for i in first:

            if i in li:

                merged = merged + li

                break

        else:

            output.append(li)

    merged = merged + first

    results.append(list(set(merged)))

    return merge(output, results)

结果如下所示：

>>> lists = [[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]]

>>> merge(lists)

[[1, 3, 5, 7], [0, 2, 4, 6]]