在给定长度为n的二进制数（0、1或无）列表的情况下，如何确定所有可能的组合？

Question

我想写一个函数，以确定给定二进制数（0、1或无）的长度为n的列表的所有可能组合

假设我的列表长度应该是3。比所需的输出应为：

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  = [...]

我尝试了以下函数，给它一个1s/0s的随机初始状态和None元素，但它没有给我想要的输出（我还尝试了其他函数，如组合-也没有想要的输出）：

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)