Python – 将列表拆分为所有可能的元组对

给定一个列表，任务是编写一个可以将其拆分为所有可能的元组对组合的Python程序。

Input : test_list = [4, 7, 5, 1, 9]

Output : [[4, 7, 5, 1, 9], [4, 7, 5, (1, 9)], [4, 7, (5, 1), 9], [4, 7, (5, 9), 1], [4, (7, 5), 1, 9], [4, (7, 5), (1, 9)], [4, (7, 1), 5, 9], [4, (7, 1), (5, 9)], [4, (7, 9), 5, 1], [4, (7, 9), (5, 1)], [(4, 7), 5, 1, 9], [(4, 7), 5, (1, 9)], [(4, 7), (5, 1), 9], [(4, 7), (5, 9), 1], [(4, 5), 7, 1, 9], [(4, 5), 7, (1, 9)], [(4, 5), (7, 1), 9], [(4, 5), (7, 9), 1], [(4, 1), 7, 5, 9], [(4, 1), 7, (5, 9)], [(4, 1), (7, 5), 9], [(4, 1), (7, 9), 5], [(4, 9), 7, 5, 1], [(4, 9), 7, (5, 1)], [(4, 9), (7, 5), 1], [(4, 9), (7, 1), 5]]

Explanation : All pairs partitions are formed.

Input : test_list = [4, 7, 5, 1]

Output : [[4, 7, 5, 1], [4, 7, (5, 1)], [4, (7, 5), 1], [4, (7, 1), 5], [(4, 7), 5, 1], [(4, 7), (5, 1)], [(4, 5), 7, 1], [(4, 5), (7, 1)], [(4, 1), 7, 5], [(4, 1), (7, 5)]]

Explanation : All pairs partitions are formed.

方法：使用切片和递归

在这里，我们从第一个元素开始创建所有对，并使用递归将多个元素通过适当的分区转换为对。

Python3

def pairings(test_list):
    if len(test_list) <= 1:
        return [test_list]
 
    # keeping 1st element and attaching every element with it
    parts = [[test_list[0]] + ele for ele in pairings(test_list[1:])]
    for idx in range(1, len(test_list)):
 
        # pairing all possible from second element
        parts.extend([[(test_list[0], test_list[idx])] +
                      ele for ele in pairings(test_list[1: idx]
                                              + test_list[idx + 1:])])
 
    return parts
 
 
# initializing list
test_list = [4, 7, 5, 1]
 
# printing original list
print("The original list is : " + str(test_list))
 
# calling util. function
res = pairings(test_list)
 
# printing result
print("Created partitions : " + str(res))

输出：

The original list is : [4, 7, 5, 1]

Created partitions : [[4, 7, 5, 1], [4, 7, (5, 1)], [4, (7, 5), 1], [4, (7, 1), 5], [(4, 7), 5, 1], [(4, 7), (5, 1)], [(4, 5), 7, 1], [(4, 5), (7, 1)], [(4, 1), 7, 5], [(4, 1), (7, 5)]]