给定长度为 n 的二进制数列表（0s 或 1s 或 None），如何确定所有可能的组合？

Question

我想编写一个函数以确定给定二进制数（0s 或 1s 或 None）的长度为 n 的列表的所有可能组合。

假设我的列表长度应为 3。所需的输出应为：

arrangement_1 = [0,0,0]
arrangement_2 = [1,0,0]
arrangement_3 = [0,1,0]
arrangement_4 = [0,0,1]
arrangement_5 = [1,1,0]
arrangement_6 = [1,0,1]
arrangement_7 = [0,1,1]
arrangement_8 = [1,1,1]
arrangement_9 = [0,0,None]
arrangement_10 = [None,0,0]
arrangement_11 = [0,None,0]
arrangement_12 = [0,None,None]
arrangement_13 = [None,0,None]
arrangement_14 = [None,None,0]
arrangement_15 = [1,1,None]
arrangement_16 = [None,1,1]
arrangement_17 = [1,None,1]
arrangement_18 = [1,None,None]
arrangement_19 = [None,1,None]
arrangement_20 = [None,None,1]
arrangement_21 = [None,1,0]
arrangement_N  = [...]

我通过给它一个 1s/0s 和 None 元素的随机初始状态来尝试以下函数，但它没有给我想要的输出（我也尝试了其他函数，如组合 - 也没有期望输出）：

def calc_permutations(list = []): # Takes list with n elements and calculates no of permutations and return dictionary of number of permutations and states

    possible_states = [] 

    for i in permutations(list,len(list)):
        possible_states.append(i)

    possible_states = {"noOfstates": len(possible_states), "states": possible_states} 
    return possible_states

Answer 1

您可以使用 itertools.product:

from itertools import product
alphabet = [0, 1, None]
for x in product(alphabet, repeat=3):
  print(x)

输出：

(0, 0, 0)
(0, 0, 1)
(0, 0, None)
(0, 1, 0)
(0, 1, 1)
(0, 1, None)
(0, None, 0)
(0, None, 1)
(0, None, None)
(1, 0, 0)
(1, 0, 1)
(1, 0, None)
(1, 1, 0)
(1, 1, 1)
(1, 1, None)
(1, None, 0)
(1, None, 1)
(1, None, None)
(None, 0, 0)
(None, 0, 1)
(None, 0, None)
(None, 1, 0)
(None, 1, 1)
(None, 1, None)
(None, None, 0)
(None, None, 1)
(None, None, None)

给定长度为 n 的二进制数列表（0s 或 1s 或 None），如何确定所有可能的组合？

How to determine all the possible combinations given a list of binary numbers (0s or 1s or None) of length n?

python

combinations

permutation