如何生成扁平列表列表的所有排列？

Question

How to generate all permutations of the flattened list of lists in Python, ... such that the order in the lists is maintained?

Example;

input;

[[1,2], [3]]

output;

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

1 is always before 2 in the permutations.

Answer 1

IIUC, you could model this as finding all topological sorts in a DAG , so I suggest you use networkx, for example:

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)))

Output

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

For the input provided, the following DiGraph is created:

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

A topological sorting imposes the constraint that 1 is always going to be present before 2 . More on all topological sortings can be found, here .

Answer 2

Interesting problem; I'm not sure whether there is a builtin in itertools for this, but this would seem to work:

Code:

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))

Output:

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

Answer 3

Starting from the permutations generator, filter by checking that all input sublists are sublists of the permutations. Sublist function from here .

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)]