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))
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)]
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)))
有趣的问题;我不确定itertools中是否有用于此的内置工具,但这似乎是可行的:
代码:
输出:
从置换生成器开始,通过检查所有输入子列表是否都是置换的子列表来进行过滤。来自here的子列表函数。你知道吗
IIUC,您可以将其建模为在DAG中查找所有拓扑排序,因此我建议您使用networkx,例如:
输出
对于提供的输入,将创建以下有向图:
一个topological sorting施加了
1
总是在2
之前出现的约束。有关所有拓扑排序的更多信息,请参见here。你知道吗相关问题 更多 >
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