finding identical rows and columns in a numpy array

Question

I have a bolean array of nxn elements and I want to check if any row is identical to another.If there are any identical rows, I want to check if the corresponding columns are also identical.

Here is an example:

A=np.array([[0, 1, 0, 0, 0, 1],
            [0, 0, 0, 1, 0, 1],
            [0, 1, 0, 0, 0, 1],
            [1, 0, 1, 0, 1, 1],
            [1, 1, 1, 0, 0, 0],
            [0, 1, 0, 1, 0, 1]])

I would like the program to find that the first and the third row are identical, and then check if the first and the third columns are also identical; which in this case they are.

Answer 1

You can use np.array_equal():

for i in range(len(A)):  # generate pairs
    for j in range(i + 1, len(A)): 
        if np.array_equal(A[i], A[j]):  # compare rows
            if np.array_equal(A[:,i], A[:,j]):  # compare columns
                print(i, j)
        else:
            pass

or using combinations():

import itertools

for pair in itertools.combinations(range(len(A)), 2):
    if np.array_equal(A[pair[0]], A[pair[1]]) and np.array_equal(A[:,pair[0]], A[:,pair[1]]):  # compare columns
        print(pair)

Answer 2

Starting with the typical way to apply np.unique to 2D arrays and have it return unique pairs:

def unique_pairs(arr):
    uview = np.ascontiguousarray(arr).view(np.dtype((np.void, arr.dtype.itemsize * arr.shape[1])))
    uvals, uidx = np.unique(uview, return_inverse=True)
    pos = np.where(np.bincount(uidx) == 2)[0]

    pairs = []
    for p in pos:
        pairs.append(np.where(uidx==p)[0])

    return np.array(pairs)

We can then do the following:

row_pairs = unique_pairs(A)
col_pairs = unique_pairs(A.T)

for pair in row_pairs:
    if np.any(np.all(pair==col_pairs, axis=1)):
        print pair

>>> [0 2]

Of course there is quite a few optimizations left to do, but the main point is using np.unique. The efficiency on this method compared to others depends heavily on how you define "small" arrays.