翻转任意维度的 numpy 数组

Question

I would like to construct a helper function that flips a multidimensional numpy array of arbitrary dimension, in all dimensions. Surprisingly, I haven't found anything online about this.

We could do something ugly like this:

d = len(X.shape)
if d == 1:
    reversed_X = X[::-1]
elif d == 2:
    reversed_X = X[::-1, ::-1]
elif d == 3:
    reversed_X = X[::-1, ::-1, ::-1]
elif d == 4:
    reversed_X = X[::-1, ::-1, ::-1, ::-1]  
# ...etc

But there has to be a better way.

I tried building a list of slice objects and using them as follows:

X[[slice(s,-1,-1) for s in X.shape]]

but this returned an empty array (!). Changing the endpoint of the slices, as in:

X[[slice(s,0,-1) for s in X.shape]]

almost works, but it misses the last index in each dimension, making the "reversed" array slightly smaller than the original.

Answer 1

Figured out the answer to my own question almost immediately after I posted it. Posting here for future people.

Slice objects can't be told to stop at -1 for some reason, even when striding backwards. But luckily you can stop at 'None', which saturates the array backwards. Applying this to the code in the original question works:

reversed_X = X[tuple([slice(None, None, -1) for _ in range(X.ndim)])]

Example:

In: X = np.reshape(np.arange(2*3), (2,3))
In: X
Out: 
array([[0, 1, 2],
       [3, 4, 5]])
In: X[tuple([slice(None, None, -1) for _ in range(X.ndim)])]
Out: 
array([[5, 4, 3],
       [2, 1, 0]])

Answer 2

numpy.flip(m, axis) (New in numpy version 1.12.0.)

So this might work:

for k in range(X.ndim):
    X = np.flip(X, k)