请问如何用python做两个数组逐位判断？

Question

比如有以下数组：
a1: 1,0,0,1,0,0,0,1
a2: 0,0,0,0,1,1,1,1
a3: 0,1,0,1,0,1,0,0
a4: 1,0,1,1,1,1,0,0
a5: .......
抓取三个数组进行判断，
if ((a1第一位or a2第一位 or a3第一位=1 )and (a1第二位 or a2 第二位 or a3第二位=1）and....
直到判断完所有位数为止，所有位都有了1的话就输出当前这三个数组，已输出的数组不参与之后的判断。

Answer 1

不知道抓取3个数组是任意3个，比如123 124 125 134 135 235...，还是顺序联系的3个，123 456 789这样抓。
我是按照前者写的，如果不是，请说明

# -*- coding: UTF-8 -*-
from itertools import combinations

a1=[ 1,0,0,1,0,0,0,1]
a2=[ 0,0,0,0,1,1,1,1]
a3=[ 0,1,0,1,0,1,0,0]
a4=[ 1,0,1,1,1,1,0,0]
a5=[ 1,1,1,1,1,1,1,0]
a6=[ 0,0,0,0,0,0,0,1]
a=[a1,a2,a3,a4,a5,a6]

al = list(combinations(a,3))
for i in al:
    flag = True
    for j in range(len(i[0])):
        if (i[0][j] + i[1][j] + i[2][j] == 0):
            flag = False
            break
    if flag:
        print(i)    

([1, 0, 0, 1, 0, 0, 0, 1], [0, 0, 0, 0, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 0])
([1, 0, 0, 1, 0, 0, 0, 1], [0, 1, 0, 1, 0, 1, 0, 0], [1, 1, 1, 1, 1, 1, 1, 0])
([1, 0, 0, 1, 0, 0, 0, 1], [1, 0, 1, 1, 1, 1, 0, 0], [1, 1, 1, 1, 1, 1, 1, 0])
([1, 0, 0, 1, 0, 0, 0, 1], [1, 1, 1, 1, 1, 1, 1, 0], [0, 0, 0, 0, 0, 0, 0, 1])
([0, 0, 0, 0, 1, 1, 1, 1], [0, 1, 0, 1, 0, 1, 0, 0], [1, 0, 1, 1, 1, 1, 0, 0])
([0, 0, 0, 0, 1, 1, 1, 1], [0, 1, 0, 1, 0, 1, 0, 0], [1, 1, 1, 1, 1, 1, 1, 0])
([0, 0, 0, 0, 1, 1, 1, 1], [1, 0, 1, 1, 1, 1, 0, 0], [1, 1, 1, 1, 1, 1, 1, 0])
([0, 0, 0, 0, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 0], [0, 0, 0, 0, 0, 0, 0, 1])
([0, 1, 0, 1, 0, 1, 0, 0], [1, 1, 1, 1, 1, 1, 1, 0], [0, 0, 0, 0, 0, 0, 0, 1])
([1, 0, 1, 1, 1, 1, 0, 0], [1, 1, 1, 1, 1, 1, 1, 0], [0, 0, 0, 0, 0, 0, 0, 1])

---

只使用1次

# -*- coding: UTF-8 -*-
from itertools import combinations

a1=[ 1,0,0,1,0,0,0,1]
a2=[ 0,0,0,0,1,1,1,1]
a3=[ 0,1,0,1,0,1,0,0]
a4=[ 1,0,1,1,1,1,0,0]
a5=[ 1,1,1,1,1,1,1,0]
a6=[ 0,0,0,0,0,0,0,1]
a=[a1,a2,a3,a4,a5,a6]
al = list(combinations(list(i for i in range(0, 6)),3))
used = []
for i in al:
    flag = False
    if not(i[0] in used or i[1] in used or i[2] in used):
        flag = True
        for j in range(7):
            if (a[i[0]][j] + a[i[1]][j] + a[i[2]][j] == 0):
                flag = False
                break
    if flag:
        used.append(i[0])
        used.append(i[1])
        used.append(i[2])
        print(' '.join(map(str, a[i[0]])) + "; " + ' '.join(map(str, a[i[1]])) + "; " + ' '.join(map(str, a[i[2]])))

1 0 0 1 0 0 0 1; 0 0 0 0 1 1 1 1; 1 1 1 1 1 1 1 0