如何从给定的列表中选择所有可能的子列表对，它们的联合将成为列表？

Question

Given a list say [4,5,6,7] i want to write a function that will output all pairs like ([4,5,7],[6]) ,( [4,5] , [6,7]) length of sub-list are not fixed

i used a for loop and two new lists to append ith element to list one and append the remaining to list 2 then repeat the process i know this is a silly approach

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)   

i expect to print all possible sub-lists but always miss the non contiguous ones , i get( [4, 5 ] , [6,7] ) but never generate ([4,6],[5,7])

Answer 1

You can use itertools module to construct all such sublists:

import itertools as it

s = {4,5,6,7}
for length in range(len(s)):
    for combination in it.combinations(s, length):
        print(list(combination), list(s - set(combination)))

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

Answer 2

Here is a solution that does not remove duplicates:

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)

Output:

[[], [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], []]

Answer 3

You can make a random sublist, then get the elements on the second list that are not iun the random sublist:

def get_from_0_to_len_minus_1_elements_from_original_list():
    pass # You implement this


list1 = get_from_0_to_len_minus_1_elements_from_original_list()
list2 = [element for element in original_list if element not in list1]
print(list1, list2)

As to how you partition the original list into random chunks that is up to you to solve