我想按照行和列的位置重新排列一个数组，在Python中怎么做？

Question

在R中，我的想法是用Python语言来做，如果我在Python语言中应用M[index]，顺序与R中的结果不同。

R中的代码：

> 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" 
> 

Python中的代码：

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

您可以使用np.flipud和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')

或者，您可以使用np.rot90旋转90度，两次

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

如果你喜欢按索引来做，你可以这样做：

>>> 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')

它本质上类似于：

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