找出兩個列表 x 和 y 之間的所有配對組合，使得 y 中的所有元素都與 x 中的一個完全配對

Question

我試圖在 python3 中找到兩個列表 x 和 y 中元素的每個唯一組合，這樣對於每個組合，列表 y 中的所有元素都與列表 x 中的一個元素配對（ min(x,y) ^ max(x,y)組合）。 例如，使用以下內容：

x = ['a', 'b', 'c']
y = [1, 2, 3]

combos = get_combos(x,y)
for combo in combos:
    print(combo)

...我想編寫get_combos(x, y)使其返回 27 個組合對的列表，打印時看起來像：

[('a', 1) ('a', 2) ('a', 3)]
[('a', 1) ('a', 2) ('b', 3)]
[('a', 1) ('a', 2) ('c', 3)]
[('a', 1) ('b', 2) ('a', 3)]
[('a', 1) ('b', 2) ('b', 3)]
[('a', 1) ('b', 2) ('c', 3)]
[('a', 1) ('c', 2) ('a', 3)]
[('a', 1) ('c', 2) ('b', 3)]
[('a', 1) ('c', 2) ('c', 3)]
[('b', 1) ('a', 2) ('a', 3)]
[('b', 1) ('a', 2) ('b', 3)]
[('b', 1) ('a', 2) ('c', 3)]
[('b', 1) ('b', 2) ('a', 3)]
[('b', 1) ('b', 2) ('b', 3)]
[('b', 1) ('b', 2) ('c', 3)]
[('b', 1) ('c', 2) ('a', 3)]
[('b', 1) ('c', 2) ('b', 3)]
[('b', 1) ('c', 2) ('c', 3)]
[('c', 1) ('a', 2) ('a', 3)]
[('c', 1) ('a', 2) ('b', 3)]
[('c', 1) ('a', 2) ('c', 3)]
[('c', 1) ('b', 2) ('a', 3)]
[('c', 1) ('b', 2) ('b', 3)]
[('c', 1) ('b', 2) ('c', 3)]
[('c', 1) ('c', 2) ('a', 3)]
[('c', 1) ('c', 2) ('b', 3)]
[('c', 1) ('c', 2) ('c', 3)]

我已經看過 itertools.combinations、itertools.product 和 itertools.permutations，但它們似乎都沒有給我我正在尋找的東西。 itertools.permutations 與zip使用時讓我很接近（請參閱此答案），但結果列表是排他的，因為任一列表中的任何元素都不能在單個組合中重復（例如[('a', 1), ('a', 2), ('c', 3)]將被排除在外），這不是我想要的。 itertools 會為此工作還是需要從頭開始編寫？

Answer 1

對我來說，它看起來像是itertools.product和zip任務，我會這樣做：

import itertools
x = ['a', 'b', 'c']
y = [1, 2, 3]
for t in itertools.product(x,repeat=3):
    print(list(zip(t,y)))

輸出：

[('a', 1), ('a', 2), ('a', 3)]
[('a', 1), ('a', 2), ('b', 3)]
[('a', 1), ('a', 2), ('c', 3)]
[('a', 1), ('b', 2), ('a', 3)]
[('a', 1), ('b', 2), ('b', 3)]
[('a', 1), ('b', 2), ('c', 3)]
[('a', 1), ('c', 2), ('a', 3)]
[('a', 1), ('c', 2), ('b', 3)]
[('a', 1), ('c', 2), ('c', 3)]
[('b', 1), ('a', 2), ('a', 3)]
[('b', 1), ('a', 2), ('b', 3)]
[('b', 1), ('a', 2), ('c', 3)]
[('b', 1), ('b', 2), ('a', 3)]
[('b', 1), ('b', 2), ('b', 3)]
[('b', 1), ('b', 2), ('c', 3)]
[('b', 1), ('c', 2), ('a', 3)]
[('b', 1), ('c', 2), ('b', 3)]
[('b', 1), ('c', 2), ('c', 3)]
[('c', 1), ('a', 2), ('a', 3)]
[('c', 1), ('a', 2), ('b', 3)]
[('c', 1), ('a', 2), ('c', 3)]
[('c', 1), ('b', 2), ('a', 3)]
[('c', 1), ('b', 2), ('b', 3)]
[('c', 1), ('b', 2), ('c', 3)]
[('c', 1), ('c', 2), ('a', 3)]
[('c', 1), ('c', 2), ('b', 3)]
[('c', 1), ('c', 2), ('c', 3)]

注意itertools.product的類似里程表的性質

Answer 2

沒有導入的基本遞歸解決方案：

x = ['a', 'b', 'c']
y = [1, 2, 3]
def groups(d, c=[]):
  if len(c) == len(x):
    yield list(zip(c, y))
  else:
    for i in d:
       yield from groups(d, c+[i])

print(list(groups(x)))

輸出：

[[('a', 1), ('a', 2), ('a', 3)], [('a', 1), ('a', 2), ('b', 3)], [('a', 1), ('a', 2), ('c', 3)], [('a', 1), ('b', 2), ('a', 3)], [('a', 1), ('b', 2), ('b', 3)], [('a', 1), ('b', 2), ('c', 3)], [('a', 1), ('c', 2), ('a', 3)], [('a', 1), ('c', 2), ('b', 3)], [('a', 1), ('c', 2), ('c', 3)], [('b', 1), ('a', 2), ('a', 3)], [('b', 1), ('a', 2), ('b', 3)], [('b', 1), ('a', 2), ('c', 3)], [('b', 1), ('b', 2), ('a', 3)], [('b', 1), ('b', 2), ('b', 3)], [('b', 1), ('b', 2), ('c', 3)], [('b', 1), ('c', 2), ('a', 3)], [('b', 1), ('c', 2), ('b', 3)], [('b', 1), ('c', 2), ('c', 3)], [('c', 1), ('a', 2), ('a', 3)], [('c', 1), ('a', 2), ('b', 3)], [('c', 1), ('a', 2), ('c', 3)], [('c', 1), ('b', 2), ('a', 3)], [('c', 1), ('b', 2), ('b', 3)], [('c', 1), ('b', 2), ('c', 3)], [('c', 1), ('c', 2), ('a', 3)], [('c', 1), ('c', 2), ('b', 3)], [('c', 1), ('c', 2), ('c', 3)]]

Answer 3

這是combinations_with_replacement 和排列組合的組合，產生了 27。

a = set([])
for i in combinations_with_replacement(['a','b','c'],3):
    for j in permutations(i):
        a.add(j)

assert len(a) == 27

for i in a:
    print(list(zip(i,[1,2,3])))

產量：

[('b', 1), ('b', 2), ('b', 3)]
[('a', 1), ('a', 2), ('c', 3)]
[('b', 1), ('a', 2), ('b', 3)]
[('c', 1), ('a', 2), ('c', 3)]
[('c', 1), ('b', 2), ('a', 3)]
[('c', 1), ('c', 2), ('c', 3)]
[('a', 1), ('c', 2), ('a', 3)]
[('c', 1), ('b', 2), ('c', 3)]
[('c', 1), ('a', 2), ('a', 3)]
[('a', 1), ('a', 2), ('a', 3)]
[('a', 1), ('c', 2), ('b', 3)]
[('a', 1), ('c', 2), ('c', 3)]
[('c', 1), ('c', 2), ('a', 3)]
[('c', 1), ('b', 2), ('b', 3)]
[('a', 1), ('b', 2), ('a', 3)]
[('c', 1), ('c', 2), ('b', 3)]
[('a', 1), ('a', 2), ('b', 3)]
[('b', 1), ('c', 2), ('a', 3)]
[('b', 1), ('b', 2), ('c', 3)]
[('c', 1), ('a', 2), ('b', 3)]
[('b', 1), ('a', 2), ('c', 3)]
[('b', 1), ('c', 2), ('c', 3)]
[('b', 1), ('a', 2), ('a', 3)]
[('a', 1), ('b', 2), ('c', 3)]
[('a', 1), ('b', 2), ('b', 3)]
[('b', 1), ('b', 2), ('a', 3)]
[('b', 1), ('c', 2), ('b', 3)]

Answer 4

嘗試這個

import itertools
L = ['a','b','c']
P = list( itertools.product(L, repeat=3) )
[ [(x,1),(y,2),(z,3)] for x,y,z in P ]