Question

如何在Python中生成扁平化列表列表的所有排列，...从而保持列表中的顺序？

示例；

输入;

[[1,2], [3]]

输出；

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

在排列中1始终在2之前。

Answer 1

有趣的问题；我不确定itertools中是否有内置功能，但这似乎可行：

代码：

l = [[1,2], [3]]

# Create list containing indexes of sublists by the number of elements in that
# sublist - in this case [0, 0, 1]
l2 = [y for i, x in enumerate(l) for y in [i]*len(x)]

rv = []

# For every unique permutation of l2:
for x in set(itertools.permutations(l2)):
    l = [[1,2], [3]]
    perm = []
    # Create a permutation from l popping the first item of a sublist when
    # we come across that sublist's index
    for i in x:
        perm.append(l[i].pop(0))
    rv.append(tuple(perm))

输出：

>>> rv
[(3, 1, 2), (1, 3, 2), (1, 2, 3)]

Answer 2

IIUC，您可以将其建模为在DAG中查找所有拓扑类别，因此我建议您使用networkx，例如：

import itertools
import networkx as nx

data = [[1,2], [3]]
edges = [edge for ls in data for edge in zip(ls, ls[1:])]

# this creates a graph from the edges (e.g. [1, 2])
dg = nx.DiGraph(edges)

# add all the posible nodes (e.g. {1, 2, 3})
dg.add_nodes_from(set(itertools.chain.from_iterable(data)))

print(list(nx.all_topological_sorts(dg)))

输出

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

对于提供的输入，将创建以下DiGraph：

Nodes: [1, 2, 3], Edges: [(1, 2)]

topological sorting施加了一个约束，即1总是要出现在2之前。可以在here上找到更多关于所有拓扑排序的信息。

Answer 3

从排列生成器开始，通过检查所有输入子列表是否是排列的子列表进行过滤。来自here的子列表功能。

l = [[1,2], [3]]

def sublist(lst1, lst2):
   ls1 = [element for element in lst1 if element in lst2]
   ls2 = [element for element in lst2 if element in lst1]
   return ls1 == ls2

[perm for perm in itertools.permutations(itertools.chain.from_iterable(l))
 if all(sublist(l_el, perm) for l_el in l)]

[(1, 2, 3), (1, 3, 2), (3, 1, 2)]

如何生成扁平化列表列表的所有排列？

3 个答案:

代码：

输出：