用Numpy重新排列矩阵元素

Question

我有一个NumPy矩阵,我已经简化为例证:

       a  b  c  d  e  f 
A =  [[0, 1, 2, 3, 4, 5],
 b    [1, 0, 3, 4, 5, 6],
 c    [2, 3, 0, 5, 6, 7],
 d    [3, 4, 5, 0, 7, 8],
 e    [4, 5, 6, 7, 0, 9],
 f    [5, 6, 7, 8, 9, 0]]

其中"十字路口"的数字很重要,但他们的顺序不对.我想重新排列行和列,以便新的顺序是[a,d,b,e,c,f]但是这个我称之为"交集"的值是相同的.

下面我开始按照我想要的方式转换矩阵.填充"e"行包括查看上面的交叉点(e,a)(= 4),然后(e,d)(= 7),然后(e,b)(= 5),(e,e) ,(e,c)和(e,f)

       a  d  b  e  c  f
A1=  [[0, 3, 1, 4, 2, 5],
 d    [3, 0, 4, 7, 5, 8],
 b    [1, 4, 0, 5, 3, 6],  
 e    [4, 7, 5, 

任何人都可以建议如何以这种方式重新安排我的矩阵？

Answer 1

编辑:我偶然发现了一个使用高级索引的NumPy解决方案:

#                 a  b  c  d  e  f
A = numpy.array([[0, 1, 2, 3, 4, 5],
                 [1, 0, 3, 4, 5, 6],
                 [2, 3, 0, 5, 6, 7],
                 [3, 4, 5, 0, 7, 8],
                 [4, 5, 6, 7, 0, 9],
                 [5, 6, 7, 8, 9, 0]])

#            a  d  b  e  c  f
new_order = [0, 3, 1, 4, 2, 5]
A1 = A[:, new_order][new_order]

这是一个纯Python解决方案,可以转移到NumPy:

#     a  b  c  d  e  f
A = [[0, 1, 2, 3, 4, 5],
     [1, 0, 3, 4, 5, 6],
     [2, 3, 0, 5, 6, 7],
     [3, 4, 5, 0, 7, 8],
     [4, 5, 6, 7, 0, 9],
     [5, 6, 7, 8, 9, 0]]

#            a  d  b  e  c  f
new_order = [0, 3, 1, 4, 2, 5]    # maps previous index to new index
A1 = [[A[i][j] for j in new_order] for i in new_order]

结果:

>>> pprint.pprint(A1)
[[0, 3, 1, 4, 2, 5],
 [3, 0, 4, 7, 5, 8],
 [1, 4, 0, 5, 3, 6],
 [4, 7, 5, 0, 6, 9],
 [2, 5, 3, 6, 0, 7],
 [5, 8, 6, 9, 7, 0]]

这是一个A适当修改的版本:

A[:] = [A[i] for i in new_order]
for row in A:
    row[:] = [row[i] for i in new_order]