Question

我想在排列和组合之间产生一些元组列表。例如，如果我有列表

list_of_tuples = [(1,20), (1,21), (2,18), (2,19)]

我想创造所有可能的＆＃34;组合＆＃34; 3个元组，以便我希望列表包含结果[(1,20), (1,20), (1,20)]，但我认为[(1,20), (1,20), (1,21)]与[(1,20), (1,21), (1,20)]和[(1,21), (1,20), (1,20)]相同，并且只想保留其中一个（与哪一个无关。

换句话说，如果＆＃34;组合＆＃34;包含与另一个＆＃34;组合相同的元组＆＃34;，我不想保留另一个元组。

我尝试过像

这样的事情

list_of_lists = [list_of_tuples]*3
results = list(itertools.product(*list_of_lists))
results = set(results)

但是使用set()我会丢失[(1,20), (1,20), (1,20)]以及所有其他具有相同元组三次的结果。

Answer 1

使用itertools.combinations_with_replacement，它应该产生你所描述的内容：

>>> from itertools import combinations_with_replacement
>>> list_of_tuples = [(1,20), (1,21), (2,18), (2,19)]
>>> list(combinations_with_replacement(list_of_tuples, 3))
[((1, 20), (1, 20), (1, 20)),
 ((1, 20), (1, 20), (1, 21)),
 ((1, 20), (1, 20), (2, 18)),
 ((1, 20), (1, 20), (2, 19)),
 ((1, 20), (1, 21), (1, 21)),
 ((1, 20), (1, 21), (2, 18)),
 ((1, 20), (1, 21), (2, 19)),
 ((1, 20), (2, 18), (2, 18)),
 ((1, 20), (2, 18), (2, 19)),
 ((1, 20), (2, 19), (2, 19)),
 ((1, 21), (1, 21), (1, 21)),
 ((1, 21), (1, 21), (2, 18)),
 ((1, 21), (1, 21), (2, 19)),
 ((1, 21), (2, 18), (2, 18)),
 ((1, 21), (2, 18), (2, 19)),
 ((1, 21), (2, 19), (2, 19)),
 ((2, 18), (2, 18), (2, 18)),
 ((2, 18), (2, 18), (2, 19)),
 ((2, 18), (2, 19), (2, 19)),
 ((2, 19), (2, 19), (2, 19))]

Answer 2

你也可以试试这个：

from itertools import product
from collections import Counter

list_of_tuples = [(1,20), (1,21), (2,18), (2,19)]

list_of_lists = [list_of_tuples] * 3

seen = set()
unique = []

for prod in product(*list_of_lists):
    curr = frozenset(Counter(prod).items())

    if curr not in seen:
        seen.add(curr)
        unique.append(prod)

print(unique)

哪个输出：

[((1, 20), (1, 20), (1, 20)), 
 ((1, 20), (1, 20), (1, 21)), 
 ((1, 20), (1, 20), (2, 18)), 
 ((1, 20), (1, 20), (2, 19)), 
 ((1, 20), (1, 21), (1, 21)), 
 ((1, 20), (1, 21), (2, 18)), 
 ((1, 20), (1, 21), (2, 19)), 
 ((1, 20), (2, 18), (2, 18)), 
 ((1, 20), (2, 18), (2, 19)), 
 ((1, 20), (2, 19), (2, 19)), 
 ((1, 21), (1, 21), (1, 21)), 
 ((1, 21), (1, 21), (2, 18)), 
 ((1, 21), (1, 21), (2, 19)), 
 ((1, 21), (2, 18), (2, 18)), 
 ((1, 21), (2, 18), (2, 19)), 
 ((1, 21), (2, 19), (2, 19)), 
 ((2, 18), (2, 18), (2, 18)), 
 ((2, 18), (2, 18), (2, 19)), 
 ((2, 18), (2, 19), (2, 19)), 
 ((2, 19), (2, 19), (2, 19))]

Answer 3

你看起来像这样吗？

list_of_tuples = [(1,20), (1,21), (2,18), (2,19)]

import itertools

data=[]

for i in itertools.product(list_of_tuples,repeat=3):
    data.append(i)

dict_1=[]

for i in data:
    if sorted(i,key=lambda x:x[1]) not in dict_1:
        dict_1.append(sorted(i,key=lambda x:x[1]))

print(dict_1)

输出：

[[(1, 20), (1, 20), (1, 20)], [(1, 20), (1, 20), (1, 21)], [(2, 18), (1, 20), (1, 20)], [(2, 19), (1, 20), (1, 20)], [(1, 20), (1, 21), (1, 21)], [(2, 18), (1, 20), (1, 21)], [(2, 19), (1, 20), (1, 21)], [(2, 18), (2, 18), (1, 20)], [(2, 18), (2, 19), (1, 20)], [(2, 19), (2, 19), (1, 20)], [(1, 21), (1, 21), (1, 21)], [(2, 18), (1, 21), (1, 21)], [(2, 19), (1, 21), (1, 21)], [(2, 18), (2, 18), (1, 21)], [(2, 18), (2, 19), (1, 21)], [(2, 19), (2, 19), (1, 21)], [(2, 18), (2, 18), (2, 18)], [(2, 18), (2, 18), (2, 19)], [(2, 18), (2, 19), (2, 19)], [(2, 19), (2, 19), (2, 19)]]

元组列表的唯一排列

3 个答案: