我想根据行和列的位置重新排列一个数组，在 Python 怎么办？

Question

In R I manage to do it, the idea is to do it in Python, if I apply M[index] in Python, the order is different from the result in R.

Code in R:

> M = matrix(c("a",0,"z",
+               0,0,"b",
+               "c","y",0), nrow = 3, byrow = TRUE)
> M
     [,1] [,2] [,3]
[1,] "a"  "0"  "z" 
[2,] "0"  "0"  "b" 
[3,] "c"  "y"  "0" 
> 
> index = c(3,2,1)
> 
> M[index,index]
     [,1] [,2] [,3]
[1,] "0"  "y"  "c" 
[2,] "b"  "0"  "0" 
[3,] "z"  "0"  "a" 
> 

Code in Python:

M = np.array([["a",0,"z"],
                  [0,0,"b"],
                  ["c","y",0]])
index = [2,1,0]

print(M[index])
array([['c', 'y', '0'],
       ['0', '0', 'b'],
       ['a', '0', 'z']], dtype='<U1')

Answer 1

You can use np.flipud and np.fliplr :

>>> M = np.array([["a",0,"z"],
                  [0,0,"b"],
                  ["c","y",0]])
>>> np.fliplr(np.flipud(M))
array([['0', 'y', 'c'],
       ['b', '0', '0'],
       ['z', '0', 'a']], dtype='<U1')

Or you can rotate by 90 degrees, twice, with np.rot90 :

>>> np.rot90(M, 2)
array([['0', 'y', 'c'],
       ['b', '0', '0'],
       ['z', '0', 'a']], dtype='<U1')

If you prefer to do it by indices, you can do:

>>> index = [2,1,0]
>>> M[index, ::-1]
array([['0', 'y', 'c'],
       ['b', '0', '0'],
       ['z', '0', 'a']], dtype='<U1')
# or,
>>> M[::-1, index]
array([['0', 'y', 'c'],
       ['b', '0', '0'],
       ['z', '0', 'a']], dtype='<U1')

Which is, essentially, similar to:

>>> M[::-1, ::-1]
array([['0', 'y', 'c'],
       ['b', '0', '0'],
       ['z', '0', 'a']], dtype='<U1')