python - 如何从给定列表中选择所有可能的子列表对，其并集将是列表？

Question

给定一个列表[4,5,6,7]，我想编写一个输出的函数所有对如 ([4,5,7],[6]) ,( [4,5] , [6,7])子列表的长度不固定

我使用了一个 for 循环和两个新列表将第 i 个元素附加到列表一个并将剩余的附加到列表 2 然后重复该过程我知道这是一个愚蠢的方法

a = [4,5,6,7]
for x in range(0,len(a)):
     b = []
     c = []
     b.append(a[x])
     for y in range(0,len(a)):
          if y!=x:
               c.append(a[y])
     print(b,c)
for x in range(0,len(a)-1):
     b = []
     c = []
     b.append(a[x])
     b.append(a[x+1])
     for y in range(0,len(a)):
          if y!=x and y!=x+1:
               c.append(a[y])
     print(b,c)   

我希望打印所有可能的子列表，但总是错过不连续的子列表，我得到( [4, 5 ] , [6,7] ) 但从未生成([4,6],[5,7])

Answer 1

这是一个不删除重复项的解决方案:

from itertools import combinations

a = [4,4,5,6,7]
result = []

for x in range(len(a) + 1):
    for left in combinations(a, x):
        right = a.copy()
        for ele in left:
            right.remove(ele)
        result.append([list(left), right])

for res in result:
    print(res)

输出:

[[], [4, 4, 5, 6, 7]]
[[4], [4, 5, 6, 7]]
[[4], [4, 5, 6, 7]]
[[5], [4, 4, 6, 7]]
[[6], [4, 4, 5, 7]]
[[7], [4, 4, 5, 6]]
[[4, 4], [5, 6, 7]]
[[4, 5], [4, 6, 7]]
[[4, 6], [4, 5, 7]]
[[4, 7], [4, 5, 6]]
[[4, 5], [4, 6, 7]]
[[4, 6], [4, 5, 7]]
[[4, 7], [4, 5, 6]]
[[5, 6], [4, 4, 7]]
[[5, 7], [4, 4, 6]]
[[6, 7], [4, 4, 5]]
[[4, 4, 5], [6, 7]]
[[4, 4, 6], [5, 7]]
[[4, 4, 7], [5, 6]]
[[4, 5, 6], [4, 7]]
[[4, 5, 7], [4, 6]]
[[4, 6, 7], [4, 5]]
[[4, 5, 6], [4, 7]]
[[4, 5, 7], [4, 6]]
[[4, 6, 7], [4, 5]]
[[5, 6, 7], [4, 4]]
[[4, 4, 5, 6], [7]]
[[4, 4, 5, 7], [6]]
[[4, 4, 6, 7], [5]]
[[4, 5, 6, 7], [4]]
[[4, 5, 6, 7], [4]]
[[4, 4, 5, 6, 7], []]