Question

我想从数组中生成组合（总是4个元素）：

arr = ["1","2","3","4"]

我目前正在使用：

itertools.combinations(arr, 2)

这会让我回复：

 ("1","2"),("1","3"),("1","4"),("2","3"),("2","4"),("3","4")

我想做的是仅生成依赖组合：

 [("1","2"),("3","4")]
 [("1","3"),("2","4")]
 [("1","4"),("2","3")]

意思是如果我有字符串"1"和"2"的组合，则只剩下其他组合"3"和"4"。我当前的解决方案仅使用itertools，因为每次计算都进行了两次（"1","2"对"3","4"一次（因此在这种情况下，对"3","4"是单独计算的;迭代在arr上再次遗漏"1"和"2"），另一次"3","4"与"1","2"配对。

我相信应该有更优雅的方法来解决这个问题。

Answer 1

为了好玩，这里有一个适用于任意数量偶数元素的解决方案，假设您希望每个依赖组合分成两个偶数组：

from itertools import combinations

arr = [1, 2, 3, 4, 5, 6]
group_size = len(arr) // 2

tuples = list(combinations(arr, group_size))
dependent_combinations = [
  [tuples[i], tuples[-i-1]]
  for i in range(len(tuples) // 2)
]
print(dependent_combinations)

输出：

[
    [(1, 2, 3), (4, 5, 6)],
    [(1, 2, 4), (3, 5, 6)],
    [(1, 2, 5), (3, 4, 6)],
    [(1, 2, 6), (3, 4, 5)],
    [(1, 3, 4), (2, 5, 6)],
    [(1, 3, 5), (2, 4, 6)],
    [(1, 3, 6), (2, 4, 5)],
    [(1, 4, 5), (2, 3, 6)],
    [(1, 4, 6), (2, 3, 5)],
    [(1, 5, 6), (2, 3, 4)]
]

Answer 2

如果数组总是包含4个元素（如您所述），则可以轻松构建所有选项：

arr = ["1","2","3","4"]
for i in range(1,4):
    print([arr[0], arr[i]], arr[1:i] + arr[i+1:])

#output: 
['1', '2'] ['3', '4']
['1', '3'] ['2', '4']
['1', '4'] ['2', '3']

从数组

2 个答案: