如何获取列表的所有排序，使列表等于另一个列表？

Question

I have lists A and B, which can have duplicates, for example:

A = ['x', 'x', 7]
B = [7, 'x', 'x']

Now I want all index permutations that permute list B into list A:

[1, 2, 0]    # because [B[1], B[2], B[0]] == A
[2, 1, 0]    # because [B[2], B[1], B[0]] == A

Is there are way to achieve this without iterating over all possible permutations? I already use

import itertools
for p in itertools.permutations(range(len(B))):
    if A == permute(B,p):

to iterate over all possible permutations and check for the ones I want, but I want to have the right ones faster.

Answer 1

You should decompose your problem in two :

• first find a particular permutation sigma_0 that maps B onto A
• find the set S_B of all the permutations that map B onto itself

Then the set you are looking after is just {sigma_0 \\circ \\sigma, sigma \\in S_B} .

Now the question becomes : how to we determine S_B ? To do this, you can just observe that if you write the set {0,1,2,..,n} (with n=2 in your case) as A_1 \\cup .. A_k , where each A_i corresponds to the indices in B that correspond to the i-th element (in your case, you would have A_1 = {1,2} and A_2 = {0} ), then each element of S_B can be written in a unique manner as a product tau_1 \\circ .. tau_k where each tau_i is a permutation that acts on A_i .

So, in your case S_B = {id, (1,2)} and you can take sigma_0 = (0,2) . It follows that the set your are after is {(0,2), (2,0,1)} .

Answer 2

Here's one way to do it. My perms function generates all valid permutations. First I collect the indexes for each element in B, then I recursively build and yield the permutations by always picking one of the still available indexes for each item in A until a permutation is ready to be yielded.

from collections import defaultdict

def perms(A, B):
    indexes = defaultdict(set)
    for i, e in enumerate(B):
        indexes[e].add(i)
    def find(perm):
        k = len(perm)
        if k == len(A):
            yield perm
            return
        I = indexes[A[k]]
        for i in list(I):
            I.remove(i)
            yield from find(perm + (i,))
            I.add(i)
    yield from find(())

Usage:

A = ['x', 'x', 7]
B = [7, 'x', 'x']

for perm in perms(A, B):
    print(perm)

Output:

(1, 2, 0)
(2, 1, 0)

Answer 3

You could do this in three steps. First, create a dictionary from list B that has items as keys and indexes as values. In your case, the dictionary for your list B:

B = [7, 'x', 'x']

  7 - [0]
'x' - [1, 2]

Then, go through your list A and build an index that maps List A indexes to their corresponding List B indexes. That is:

A = ['x', 'x', 7]

0 - [1, 2]
1 - [1, 2]
2 - [0]

From that, you can generate all of the valid mappings. It should be easy enough to write code that, given the index above, will generate [1, 2, 0] and [2, 1, 0] .

Answer 4

Here's something in Python. I used strings for hashing since I'm not familiar with how to pass sets and arrays as recursive arguments in Python, although they might be more efficient.

A = ['x', 'x', 7, 'y', 8, 8]
B = [7, 'x', 8, 'x', 8, 'y']
H = {}

# record the indexes of elements in B
for i in xrange(0,len(B)):
  if B[i] in H:
    H[ B[i] ].append(str(i)) 
  else:
    H[ B[i] ] = [str(i)]

# build permutations in the order that the elements are encountered in A
def perms(perm,i,visited,l):
  if i==l:
    print perm
    return
  for j in H[ A[i] ]:
    if j not in visited:
      perms(perm + j,i + 1,visited + j,l)

perms("",0,"",len(A)) 

Output:

130524
130542
310524
310542