Question

问题：如何在下面实施double_permutations(s)？

>>> s = [('a', 'b'), ('c', 'd'), ('e', 'f')]
>>> for answer in double_permutation(s):
...     print(answer)   # in some order
[('a', 'b'), ('c', 'd'), ('e', 'f')]
[('a', 'b'), ('d', 'c'), ('e', 'f')]
[('a', 'b'), ('c', 'd'), ('f', 'e')]
[('a', 'b'), ('d', 'c'), ('f', 'e')]
[('a', 'b'), ('e', 'f'), ('c', 'd')]
[('a', 'b'), ('f', 'e'), ('c', 'd')]
[('a', 'b'), ('e', 'f'), ('d', 'c')]
[('a', 'b'), ('f', 'e'), ('d', 'c')]

我尝试了什么（一旦外部列表超过3个元素就会崩溃）

from itertools import permutations

def double_permutation(l):

    def double_permutation_recur(s, r):
        if not r:
            yield s
        else:
            for permutation in permutations(r):
                s1 = s + [permutation[0]]
                s2 = s + [(permutation[0][1], permutation[0][0])]
                for perm1 in double_permutation_recur(s1, permutation[1:]):
                    yield perm1
                for perm2 in double_permutation_recur(s2, permutation[1:]):
                    yield perm2

    return double_permutation_recur([l[0]], l[1:])

对于长度为double_factorial(n-1)的列表，这应该会产生n个答案。这可以通过n = 3进行，但会在n = 4处分解（产生96而不是48个答案）。

Answer 1

您可以从itertools模块

中的基元构建它

import itertools
s = [('a', 'b'), ('c', 'd'), ('e', 'f')]

这是你所描述的吗？

def permute(it):
    return itertools.product(*(itertools.permutations(i) for i in it))

>>> for i in permute(s):
...     print i
(('a', 'b'), ('c', 'd'), ('e', 'f'))
(('a', 'b'), ('c', 'd'), ('f', 'e'))
(('a', 'b'), ('d', 'c'), ('e', 'f'))
(('a', 'b'), ('d', 'c'), ('f', 'e'))
(('b', 'a'), ('c', 'd'), ('e', 'f'))
(('b', 'a'), ('c', 'd'), ('f', 'e'))
(('b', 'a'), ('d', 'c'), ('e', 'f'))
(('b', 'a'), ('d', 'c'), ('f', 'e'))

或者你想要：

def permute2(it):
    return itertools.chain.from_iterable(
        permute(p)
        for p in itertools.permutations(it)
    )

>>> for i in permute2(s):
...      print i    
(('a', 'b'), ('c', 'd'), ('e', 'f'))
(('a', 'b'), ('c', 'd'), ('f', 'e'))
(('a', 'b'), ('d', 'c'), ('e', 'f'))
(('a', 'b'), ('d', 'c'), ('f', 'e'))
(('b', 'a'), ('c', 'd'), ('e', 'f'))
(('b', 'a'), ('c', 'd'), ('f', 'e'))
(('b', 'a'), ('d', 'c'), ('e', 'f'))
(('b', 'a'), ('d', 'c'), ('f', 'e'))
(('a', 'b'), ('e', 'f'), ('c', 'd'))
(('a', 'b'), ('e', 'f'), ('d', 'c'))
(('a', 'b'), ('f', 'e'), ('c', 'd'))
(('a', 'b'), ('f', 'e'), ('d', 'c'))
(('b', 'a'), ('e', 'f'), ('c', 'd'))
(('b', 'a'), ('e', 'f'), ('d', 'c'))
(('b', 'a'), ('f', 'e'), ('c', 'd'))
(('b', 'a'), ('f', 'e'), ('d', 'c'))
(('c', 'd'), ('a', 'b'), ('e', 'f'))
(('c', 'd'), ('a', 'b'), ('f', 'e'))
(('c', 'd'), ('b', 'a'), ('e', 'f'))
(('c', 'd'), ('b', 'a'), ('f', 'e'))
(('d', 'c'), ('a', 'b'), ('e', 'f'))
(('d', 'c'), ('a', 'b'), ('f', 'e'))
(('d', 'c'), ('b', 'a'), ('e', 'f'))
(('d', 'c'), ('b', 'a'), ('f', 'e'))
(('c', 'd'), ('e', 'f'), ('a', 'b'))
(('c', 'd'), ('e', 'f'), ('b', 'a'))
(('c', 'd'), ('f', 'e'), ('a', 'b'))
(('c', 'd'), ('f', 'e'), ('b', 'a'))
(('d', 'c'), ('e', 'f'), ('a', 'b'))
(('d', 'c'), ('e', 'f'), ('b', 'a'))
(('d', 'c'), ('f', 'e'), ('a', 'b'))
(('d', 'c'), ('f', 'e'), ('b', 'a'))
(('e', 'f'), ('a', 'b'), ('c', 'd'))
(('e', 'f'), ('a', 'b'), ('d', 'c'))
(('e', 'f'), ('b', 'a'), ('c', 'd'))
(('e', 'f'), ('b', 'a'), ('d', 'c'))
(('f', 'e'), ('a', 'b'), ('c', 'd'))
(('f', 'e'), ('a', 'b'), ('d', 'c'))
(('f', 'e'), ('b', 'a'), ('c', 'd'))
(('f', 'e'), ('b', 'a'), ('d', 'c'))
(('e', 'f'), ('c', 'd'), ('a', 'b'))
(('e', 'f'), ('c', 'd'), ('b', 'a'))
(('e', 'f'), ('d', 'c'), ('a', 'b'))
(('e', 'f'), ('d', 'c'), ('b', 'a'))
(('f', 'e'), ('c', 'd'), ('a', 'b'))
(('f', 'e'), ('c', 'd'), ('b', 'a'))
(('f', 'e'), ('d', 'c'), ('a', 'b'))
(('f', 'e'), ('d', 'c'), ('b', 'a'))

或者“锚定”第一个元素：

def permute3(s):
    return s[:1] + list(p) for p in permute2(s[1:])

>>> for i in permute3(s):
...     print i    
[('a', 'b'), ('c', 'd'), ('e', 'f')]
[('a', 'b'), ('c', 'd'), ('f', 'e')]
[('a', 'b'), ('d', 'c'), ('e', 'f')]
[('a', 'b'), ('d', 'c'), ('f', 'e')]
[('a', 'b'), ('e', 'f'), ('c', 'd')]
[('a', 'b'), ('e', 'f'), ('d', 'c')]
[('a', 'b'), ('f', 'e'), ('c', 'd')]
[('a', 'b'), ('f', 'e'), ('d', 'c')]

序列和子序列的Python排列

1 个答案: