Question

我想知道如何从列表中获得所有可能的组合。

list1 = [1,5,3]

这应该返回：

ans = [1,5,3,15,13,51,53,31,35,153,135,315,351,531,513]

感谢回复
亚当

Answer 1

您需要permutations，而不是combinations：

from itertools import permutations

list1 = [1,5,3]
n = len(list1)

print([int("".join(map(str, ints))) for i in range(n)
       for ints in permutations(list1, i + 1)])
# [1, 5, 3, 15, 13, 51, 53, 31, 35, 153, 135, 513, 531, 315, 351]

Answer 2

哟可以这样做：

import itertools
import operator
import numpy as np
a=[1,2,3]
[list(i) for i in reduce(operator.add,[list(itertools.combinations(a,i)) for i in range(1,len(a)+1)])]

然后结果是

[[1], [2], [3], [1, 2], [1, 3], [2, 3], [1, 2, 3]]

如果您希望它具有您在问题中列出的确切格式，则可以执行

import itertools
import operator
import numpy as np
a=[1,2,3]
a=[str(i) for i in a]
[int(''.join(list(i))) for i in reduce(operator.add,[list(itertools.combinations(a,i)) for i in range(1,len(a)+1)])]

结果是

[1, 2, 3, 12, 13, 23, 123]

Answer 3

蛮力：

import random

list1 = [1,5,3]
brute = 1000

combos = []
for i in range(brute):
    random.shuffle(list1)
    z = ''.join(str(x) for x in list1)
    for i in range(len(list1)):
        combos.append(int(z[i:]))
combos = sorted(list(set(combos)))
print combos




>>>[1, 3, 5, 13, 15, 31, 35, 51, 53, 135, 153, 315, 351, 513, 531]

使用列表的Python组合

3 个答案: