Numpy Unique 2D Sub-array

I have 3D numpy array and I want only unique 2D-sub-arrays. Input: [[[ 1 2] [ 3 4]] [[ 5 6] [ 7 8]] [[ 9 10] [11 12]] [[ 5 6] [ 7 8]]] Output: [[[ 1 2] [ 3

Solution 1:

Using @Jaime's post, to solve our case of finding unique 2D subarrays, I came up with this solution that basically adds a reshaping to the view step -

def unique2D_subarray(a):
    dtype1 = np.dtype((np.void, a.dtype.itemsize * np.prod(a.shape[1:])))
    b = np.ascontiguousarray(a.reshape(a.shape[0],-1)).view(dtype1)
    return a[np.unique(b, return_index=1)[1]]

Sample run -

In [62]: a
Out[62]: 
array([[[ 1,  2],
        [ 3,  4]],

       [[ 5,  6],
        [ 7,  8]],

       [[ 9, 10],
        [11, 12]],

       [[ 5,  6],
        [ 7,  8]]])

In [63]: unique2D_subarray(a)
Out[63]: 
array([[[ 1,  2],
        [ 3,  4]],

       [[ 5,  6],
        [ 7,  8]],

       [[ 9, 10],
        [11, 12]]])

Solution 2:

The numpy_indexed package (disclaimer: I am its author) is designed to do operations such as these in an efficient and vectorized manner:

import numpy_indexed as npi
npi.unique(a)

Solution 3:

One solution would be to use a set to keep track of which sub arrays you have seen:

seen = set([])
new_a = []

for j in a:
    f = tuple(list(j.flatten()))
    if f not in seen:
        new_a.append(j)
        seen.add(f)

print np.array(new_a)

Or using numpy only:

print np.unique(a).reshape((len(unique) / 4, 2, 2))

>>> [[[ 1  2]
      [ 3  4]]

     [[ 5  6]
      [ 7  8]]

     [[ 9 10]
      [11 12]]]