具有唯一值的排列排列根據(jù)元素的位置而不是其值來(lái)生成元素被視為唯一的位置。所以基本上我想避免這樣的重復(fù)：>>> list(itertools.permutations([1, 1, 1]))[(1, 1, 1), (1, 1, 1), (1, 1, 1), (1, 1, 1), (1, 1, 1), (1, 1, 1)]事后過濾是不可能的，因?yàn)樵谖业那闆r下排列的數(shù)量太大。有誰(shuí)知道一個(gè)合適的算法嗎？非常感謝!編輯：我主要想要的是：x = itertools.product((0, 1, 'x'), repeat=X)x = sorted(x, key=functools.partial(count_elements, elem='x'))這是不可能的，因?yàn)閟orted創(chuàng)建一個(gè)列表，則itertools.Products的輸出太大。對(duì)不起，我應(yīng)該描述一下實(shí)際的問題。
3 回答
陪伴而非守候
TA貢獻(xiàn)1757條經(jīng)驗(yàn) 獲得超8個(gè)贊
class unique_element: def __init__(self,value,occurrences): self.value = value self.occurrences = occurrencesdef perm_unique(elements): eset=set(elements) listunique = [unique_element(i,elements.count(i)) for i in eset] u=len(elements) return perm_unique_helper(listunique,[0]*u,u-1)def perm_unique_helper(listunique,result_list,d): if d < 0: yield tuple(result_list) else: for i in listunique: if i.occurrences > 0: result_list[d]=i.value i.occurrences-=1 for g in perm_unique_helper(listunique,result_list,d-1): yield g i.occurrences+=1a = list(perm_unique([1,1,2]))print(a)
[(2, 1, 1), (1, 2, 1), (1, 1, 2)]
def permutations_with_replacement(elements,n): return permutations_helper(elements,[0]*n,n-1)#this is generatordef permutations_helper(elements,result_list,d): if d<0: yield tuple(result_list) else: for i in elements: result_list[d]=i all_permutations = permutations_helper(elements,result_list,d-1)#this is generator for g in all_permutations: yield g
yieldfor g in perm_unique_helper(listunique,result_list,d-1): yield g
permutations_with_replacement
胡子哥哥
TA貢獻(xiàn)1825條經(jīng)驗(yàn) 獲得超6個(gè)贊
from itertools import permutationsdef unique_permutations(iterable, r=None):
previous = tuple()
for p in permutations(sorted(iterable), r):
if p > previous:
previous = p yield pfor p in unique_permutations('cabcab', 2):
print p('a', 'a')('a', 'b')('a', 'c')('b', 'a')('b', 'b')('b', 'c')('c', 'a')('c', 'b')('c', 'c')添加回答
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