如何将列表与列表列表中的公共元素合并？

Question

我正在尝试将所有列表合并到具有公共元素的列表列表中。 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

我同意@Ajax1234，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]]

Answer 2

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]]