python中所有可能的列表合并

Question

I would like to get all the possible ways to merge 2 (or ideally n) lists in python, while maintaining each list's internal order - ie I would like to find some function all_merges that would behave like this:

a = all_merges([1,2],[3,4]) 

results in:

a = [
[1,2,3,4],
[1,3,2,4],
[1,3,4,2],
[3,1,2,4],
[3,1,4,2],
[3,4,1,2]   ]

(I hope this is all of them)

I cannot find simple, 'pythonic' way to do this - please help!

=====

Note:

I am considering something like this (written to be relatively readable):

from itertools import permutations
def all_merges(list1,list2):
    a = [1]*len(list1)
    b = [2]*len(list2)
    indexes = list(set(list(permutations(a+b))))
    lists = [[]]*3
    res = indexes # just for readability, will be overwriting indexes as we go along
    for perm in indexes:
        lists[1] = copy(list1)
        lists[2] = copy(list2)
        merge = perm
        for i in range(0,len(perm)):
            merge[j] = lists[perm[i]].pop(0)
    return res

however at the stage

list(set(list(permutations(a+b))

if the combined length of the lists is so much as 15 I get a huge amount of results from

permutations(a+b)

(15!, to be precise), whereas at most I will only really have 15choose7 (= 6435) different merges.

I realise a replacement to that line is provided here, as a function: permutations with unique values but by now this is getting messy and I want to see if there is a cleaner solution to my original problem.

Answer 1

Each possible merge corresponds directly to one of (len(a) + len(b)) choose len(a) ways to pick len(a) positions for the elements of a in the final list:

import itertools
def all_merges(a, b):
    # object guaranteed not to be in either input list
    sentinel = object()
    merged_length = len(a) + len(b)
    for a_positions in itertools.combinations(xrange(merged_length), len(a)):
        merged = [sentinel] * merged_length

        # Place the elements of a in their chosen positions.
        for pos, a_elem in zip(a_positions, a):
            merged[pos] = a_elem

        # Place the elements of b in the positions not taken.
        b_iter = iter(b)
        for pos in xrange(merged_length):
            if merged[pos] is sentinel:
                merged[pos] = next(b_iter)

        yield merged

This can be extended to more lists in several ways, such as by using some technique (perhaps Algorithm L ) to go through all ways to assign positions to list elements, or by applying the 2-way merge repeatedly.