您可以itertools
模塊
import itertools
s = [('a', 'b'), ('c', 'd'), ('e', 'f')]
從原語建立這個了這是你描述?
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'))
還是你想:
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'))
或者爲 「錨」 的第一個元素:
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')]
您的例子似乎不包括'(「B」,「A」)',這使得它看起來像它的分解給我,但你不提它,你說你的代碼通過'n = 3'工作。所以我必須錯過一些東西:你能解釋一些關於建築的事嗎? – DSM
我想要將第一個項目錨定。我可以改變這個,如果我想改變呼叫到「double_permutations_recur([],l) – MikeRand