numpy中的数组索引

Question

Is there a way in numpy to retrieve all items in an array except the item of the index provided.

 x = 
 array([[[4, 2, 3],
    [2, 0, 1],
    [1, 3, 4]],

   [[2, 1, 2],
    [3, 2, 3],
    [3, 4, 2]],

   [[2, 4, 1],
    [0, 2, 2],
    [4, 0, 0]]])

and by asking for

x[not 1,:,:] 

you will get

array([[[4, 2, 3],
    [2, 0, 1],
    [1, 3, 4]],

   [[2, 4, 1],
    [0, 2, 2],
    [4, 0, 0]]])

Thanks

Answer 1

In [42]: x[np.arange(x.shape[0])!=1,:,:]
Out[42]: 
array([[[4, 2, 3],
        [2, 0, 1],
        [1, 3, 4]],

       [[2, 4, 1],
        [0, 2, 2],
        [4, 0, 0]]])

Answer 2

Have you tried this?

a[(0,2), :, :]

Instead of blacklisting what you don't want to get, you can try to whitelist what you need.

If you need to blacklist anyway, you can do something like this:

a[[i for i in range(a.shape[0]) if i != 1], :, :]

Basically you just create a list with all possible indexes ( range(a.shape[0]) ) and filter out those that you don't want to get displayed ( if i != 1 ).

Answer 3

这是一个非常通用的解决方案：

x[range(0,i)+range(i+1,x.shape[0]),:,:]