在您使用k=3的情况下，np.rot90的代码可以做到:

The code for np.rot90 does, in your case of k=3:

    # k == 3
    return fliplr(m.swapaxes(0, 1))

所以

In [789]: np.fliplr(ar.swapaxes(0, 1))
Out[789]: 
array([[-11,   0],
     ...
       [ -3,   8],
       [ -2,   9]])

所以你的

fliplr(rot90(ar, 3))

成为

 np.fliplf(np.fliplr(ar.swapaxes(0, 1)))
 # the flips cancel
 ar.swapaxes(0,1)
 # but this is just
 ar.T

因此，您的一对动作会减少以进行转置.

So your pair of actions reduce to transpose.

transpose(和swap)仅更改数组的.shape和strides属性；这是一个视图，而不是副本.

transpose (and the swap) just changes the .shape and strides attributes of the array; it is a view, not a copy.

np.fliplr还会创建一个视图，并随[:,::-1]改变步幅.

np.fliplr also creates a view, changing strides with the [:,::-1].

原始的ar:

In [818]: ar
Out[818]: 
array([[  0,   1,   2,   3,   4,   5,   6,   7,   8,   9],
       [-11, -10,  -9,  -8,  -7,  -6,  -5,  -4,  -3,  -2]])

In [819]: x=np.fliplr(np.rot90(ar,3))  # your pair of actions

In [820]: x
Out[820]: 
array([[  0, -11],
       [  1, -10],
         ...
       [  8,  -3],
       [  9,  -2]])

In [821]: x[0,1]=11

In [822]: x
Out[822]: 
array([[  0,  11],
       [  1, -10],
        ...
       [  9,  -2]])

In [823]: ar
Out[823]: 
array([[  0,   1,   2,   3,   4,   5,   6,   7,   8,   9],
       [ 11, -10,  -9,  -8,  -7,  -6,  -5,  -4,  -3,  -2]])

更改x的值将更改ar的值.尽管使用了2个功能，x仍然是ar的view.

Changing a value of x changes a value of ar. Despite the use of 2 functions, x is still a view of ar.

这两个函数不是必需的，但它们也不是那么昂贵.我们所说的是微秒v纳秒的时间. (我在Ipython中的timeit时间要小得多)

The 2 functions aren't needed, but they aren't that expensive either. We are talking microseconds v nanoseconds of time. (my timeit times in Ipython are much smaller yours)

In [824]: timeit np.fliplr(np.rot90(ar,3))
100000 loops, best of 3: 8.28 µs per loop

In [825]: timeit ar.T
1000000 loops, best of 3: 455 ns per loop