Python 列表的子列表

我想要一个函数

split_list（[1,2,4,3]）

，它生成以下7个列表：

[[1], [2,4,3]]
[[1,2], [4,3]]
[[1,2,4], [3]]
[[1,2], [4],[3]]
[[1],[2,4],[3]]
[[1],[2],[4,3]]
[[1],[2],[4],[3]]

如果先生成较小的列表-

[[1]，[2,4,3]]

的长度为2，而

[[1]，[2]，[4]，[3]]

的长度为4，因此最后生成。这就是我所能做到的：

def split_list(l):
  for i in range(1, len(l)):
    yield [l[:i],l[i:]]

for i in split_list((1,2,4,3)):
    print(i)

---

您可以使用递归的列表切片：

def split_list(d):
   for i in range(len(d)):
      if len(d[i+1:]) > 0:
        for c in split_list(d[i+1:]):
          yield [d[:i+1], *c]
      else:
        yield [d]

print(list(split_list([1,2,4,3])))

输出：

[[[1], [2], [4], [3]], [[1], [2], [4, 3]], [[1], [2, 4], [3]], [[1], [2, 4, 3]], [[1, 2], [4], [3]], [[1, 2], [4, 3]], [[1, 2, 4], [3]], [[1, 2, 4, 3]]]

[[[1], [2], [4], [3]], [[1], [2], [4, 3]], [[1], [2, 4], [3]], [[1], [2, 4, 3]], [[1, 2], [4], [3]], [[1, 2], [4, 3]], [[1, 2, 4], [3]], [[1, 2, 4, 3]]]

编辑：事件较短

def split_list(d):
  if len(d) > 1:
    return [list(filter(None, [d[:i+1], *c])) for i in range(len(d)) for c in split_list(d[i+1:])]
  return [d] if not isinstance(d, list) else [[d]]

输出：

[[[1], [2], [4], [3]], [[1], [2], [4, 3]], [[1], [2, 4], [3]], [[1], [2, 4, 3]], [[1, 2], [4], [3]], [[1, 2], [4, 3]], [[1, 2, 4], [3]], [[1, 2, 4, 3]]]

[[[1], [2], [4], [3]], [[1], [2], [4, 3]], [[1], [2, 4], [3]], [[1], [2, 4, 3]], [[1, 2], [4], [3]], [[1, 2], [4, 3]], [[1, 2, 4], [3]], [[1, 2, 4, 3]]]

---

如果将列表本身作为第一个元素不成问题，可以这样做：

def split_list(lst):
    if not lst:
        yield []

    for i in range(len(lst), 0, -1):
        for j in split_list(lst[i:]):
            yield [lst[:i]] + j


for l in split_list([1, 2, 4, 3]):
    print(l)

输出

[[1, 2, 4, 3]]
[[1, 2, 4], [3]]
[[1, 2], [4, 3]]
[[1, 2], [4], [3]]
[[1], [2, 4, 3]]
[[1], [2, 4], [3]]
[[1], [2], [4, 3]]
[[1], [2], [4], [3]]

试试这个：

def split_list(l):
    if len(l) == 1:
        yield [ l ]
        return

    for s in split_list(l[1:]):
        for n, sub in enumerate(s):
            if len((s[:n] + [[ l[0] ] + sub]  + s[n+1:])[0]) !=4:
                yield s[:n] + [[ l[0] ] + sub]  + s[n+1:]           
        yield [[l[0]]] + s

for lst in split_list([1, 2, 3, 4]):
        print lst

这个问题是递归解决的。 在产量方面：

[[1], [2, 3, 4]]
[[1, 2], [3, 4]]
[[2], [1, 3, 4]]
[[1], [2], [3, 4]]
[[1, 2, 3], [4]]
[[2, 3], [1, 4]]
[[1], [2, 3], [4]]
[[1, 3], [2, 4]]
[[3], [1, 2, 4]]
[[1], [3], [2, 4]]
[[1, 2], [3], [4]]
[[2], [1, 3], [4]]
[[2], [3], [1, 4]]
[[1], [2], [3], [4]]

---

较小的列表是什么意思？@RobS的[[1]、[2,4,3]]的长度是2，而[[1]、[2]、[4]、[3]]的长度是4。可能重复：或者：是否要将主列表排除为第一个元素？例如[[1,2,3,4]]很好！！*in*c是做什么的？打开列表？@RobS是的，没错<代码>*c是简单的解包。我认为将列表与

+

连接起来有点干净（尽管这有点主观）。