如何用python生成符合该条件的全部矩阵

Question

问题遇到的现象和发生背景

在处理一个matching问题。当有六个人时
图中第二列为一个配对可能，且对于n为6时，共有15种可能

第二列的配对对应的矩阵为

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

问题相关代码，请勿粘贴截图

运行结果及报错内容

我的解答思路和尝试过的方法

我想要达到的结果

想要编一个程序可输出全部15中配对可能
希望输出结果为

apij=np.array([[[0,1,0,0,0,0],[1,0,0,0,0,0],[0,0,0,1,0,0],[0,0,1,0,0,0],[0,0,0,0,0,1],[0,0,0,0,1,0]],\
               [[0,1,0,0,0,0],[1,0,0,0,0,0],[0,0,0,0,1,0],[0,0,0,0,0,1],[0,0,1,0,0,1],[0,0,0,1,0,0]],\
               [[0,1,0,0,0,0],[1,0,0,0,0,0],[0,0,0,0,0,1],[0,0,0,0,1,0],[0,0,0,1,0,0],[0,0,1,0,0,0]],\
               [[0,0,1,0,0,0],[0,0,0,1,0,0],[1,0,0,0,0,0],[0,1,0,0,0,0],[0,0,0,0,0,1],[0,0,0,0,1,0]],\
               [[0,0,1,0,0,0],[0,0,0,0,1,0],[1,0,0,0,0,0],[0,0,0,0,0,1],[0,1,0,0,0,0],[0,0,0,1,0,0]],\
               [[0,0,1,0,0,0],[0,0,0,0,0,1],[1,0,0,0,0,0],[0,0,0,0,1,0],[0,0,0,1,0,0],[0,1,0,0,0,0]],\
               [[0,0,0,1,0,0],[0,0,1,0,0,0],[0,1,0,0,0,0],[1,0,0,0,0,0],[0,0,0,0,0,1],[0,0,0,0,1,0]],\
               [[0,0,0,1,0,0],[0,0,0,0,1,0],[0,0,0,0,0,1],[1,0,0,0,0,0],[0,1,0,0,0,0],[0,0,1,0,0,0]],\
               [[0,0,0,1,0,0],[0,0,0,0,0,1],[0,0,0,0,1,0],[1,0,0,0,0,0],[0,0,1,0,0,0],[0,1,0,0,0,0]],\
               [[0,0,0,0,1,0],[0,0,1,0,0,0],[0,1,0,0,0,0],[0,0,0,0,0,1],[1,0,0,0,0,0],[0,0,0,1,0,0]],\
               [[0,0,0,0,1,0],[0,0,0,1,0,0],[0,0,0,0,0,1],[0,1,0,0,0,0],[1,0,0,0,0,0],[0,0,1,0,0,0]],\
               [[0,0,0,0,1,0],[0,0,0,0,0,1],[0,0,0,1,0,0],[0,0,1,0,0,0],[1,0,0,0,0,0],[0,1,0,0,0,0]],\
               [[0,0,0,0,0,1],[0,0,1,0,0,0],[0,1,0,0,0,0],[0,0,0,0,1,0],[0,0,0,1,0,0],[1,0,0,0,0,0]],\
               [[0,0,0,0,0,1],[0,0,0,1,0,0],[0,0,0,0,1,0],[0,1,0,0,0,0],[0,0,1,0,0,0],[1,0,0,0,0,0]],\
               [[0,0,0,0,0,1],[0,0,0,0,1,0],[0,0,0,1,0,0],[0,0,1,0,0,0],[0,1,0,0,0,0],[1,0,0,0,0,0]]])

Answer 1

这样应该是你要的效果吧，转成ndarray比较容易

from itertools import permutations

def code(a)->dict:
    if len(a) == 2: return {ord(a[0]):ord(a[1]),ord(a[1]):ord(a[0])}
    return {**{ord(a[0]):ord(a[1]),ord(a[1]):ord(a[0])}, **code(a[2:])}

def binary(a)->list:
    res = []
    for i in a:
        j = ord(i)-65
        temp = [0]*len(a)
        temp[j]=1
        res.append(temp)
    return res

a = 'ABCDEF'
b = permutations(a)
res = set()
for i in b:
    res.add(a.translate(code(i)))
res=sorted(list(res))
for i in res:
    print(binary(i))