在列表中找到2 ^ n -2个元素的组合

Question

I have the following list:

list1 = ['g1','g2','g3','g4']

I want to find 2^n-2 combinations where n is the total number of items in the list. For n = 4 the result should be 2^4 -2 = 14 , ie 14 combinations.

The combinations are:

[[['g1'],['g2','g3','g4']],[['g2'],['g1','g3','g4']], [['g3'],['g1','g2','g4']],['g4'],['g1','g2','g3']],[['g1','g2'],['g3','g4']],[['g1','g3'],['g2','g4']],[['g1','g4'],['g3','g4']],[['g2','g3'],['g1','g4']],
[['g2','g4'],['g1','g3']],[['g3','g4'],['g1','g2']],[['g1','g2','g3'],['g4']],[['g2','g3','g4'],['g1']],[['g3','g4','g1'],['g2']],[['g4','g1','g2'],['g3']]]

I know one approach: in first iteration take single element and put it into a list and other elements in second list: ['g1'],['g2','g3','g4'] in second iteration take 2 elements in a list and other elements in second list. ['g1','g2'],['g1','g4'] Is there any other approach ?? I'm writing this program in python. My approach is costly. Is there any library method to do this quickly.

Answer 1

Here's a functional approach using itertools

import itertools as iter

list1 = ['g1', 'g2', 'g3', 'g4']
combinations = [iter.combinations(list1, n) for n in range(1, len(list1))]
flat_combinations = iter.chain.from_iterable(combinations)
result = map(lambda x: [list(x), list(set(list1) - set(x))], flat_combinations)
# [[['g1'], ['g4', 'g3', 'g2']], [['g2'], ['g4', 'g3', 'g1']], [['g3'], ['g4', 'g2', 'g1']],...
len(result)
# 14

Answer 2

itertools.combinations(iterable, r)

Return r length subsequences of elements from the input iterable. Combinations are emitted in lexicographic sort order. So, if the input iterable is sorted, the combination tuples will be produced in sorted order.

from itertools import combinations
list1 = ['g1','g2','g3','g4']
for n in range(1,len(list1)):
    for i in combinations(list1,n):
        print(set(i), set(list1) - set(i))

out:

{'g1'} {'g2', 'g3', 'g4'}
{'g2'} {'g1', 'g3', 'g4'}
{'g3'} {'g1', 'g2', 'g4'}
{'g4'} {'g1', 'g2', 'g3'}
{'g1', 'g2'} {'g3', 'g4'}
{'g1', 'g3'} {'g2', 'g4'}
{'g1', 'g4'} {'g2', 'g3'}
{'g2', 'g3'} {'g1', 'g4'}
{'g2', 'g4'} {'g1', 'g3'}
{'g3', 'g4'} {'g1', 'g2'}
{'g1', 'g2', 'g3'} {'g4'}
{'g1', 'g2', 'g4'} {'g3'}
{'g1', 'g3', 'g4'} {'g2'}
{'g2', 'g3', 'g4'} {'g1'}

you can try this

Answer 3

I like the solution from the Chinese coder (I guess). Here's my own solution:

import itertools


def flatten(*z):
    return z


list1 = ['g1','g2','g3','g4']

sublists = []
for i in range(1, len(list1)):
    sublists.extend(itertools.combinations(list1, i))

pairs = []
for a, b in itertools.product(sublists, sublists):
    if len(a) + len(b) == len(list1) and \
    len(set(flatten(*a, *b))) == len(list1):
        pairs.append((a, b))

print(pairs, len(pairs))