Find unique rows in numpy.array

As of NumPy 1.13, one can simply choose the axis for selection of unique values in any N-dim array. To get unique rows, one can do:

unique_rows = np.unique(original_array, axis=0)

Yet another possible solution

np.vstack({tuple(row) for row in a})

Another option to the use of structured arrays is using a view of a void type that joins the whole row into a single item:

a = np.array([[1, 1, 1, 0, 0, 0],              [0, 1, 1, 1, 0, 0],              [0, 1, 1, 1, 0, 0],              [1, 1, 1, 0, 0, 0],              [1, 1, 1, 1, 1, 0]])b = np.ascontiguousarray(a).view(np.dtype((np.void, a.dtype.itemsize * a.shape[1])))_, idx = np.unique(b, return_index=True)unique_a = a[idx]>>> unique_aarray([[0, 1, 1, 1, 0, 0],       [1, 1, 1, 0, 0, 0],       [1, 1, 1, 1, 1, 0]])

EDITAdded np.ascontiguousarray following @seberg's recommendation. This will slow the method down if the array is not already contiguous.

EDITThe above can be slightly sped up, perhaps at the cost of clarity, by doing:

unique_a = np.unique(b).view(a.dtype).reshape(-1, a.shape[1])

Also, at least on my system, performance wise it is on par, or even better, than the lexsort method:

a = np.random.randint(2, size=(10000, 6))%timeit np.unique(a.view(np.dtype((np.void, a.dtype.itemsize*a.shape[1])))).view(a.dtype).reshape(-1, a.shape[1])100 loops, best of 3: 3.17 ms per loop%timeit ind = np.lexsort(a.T); a[np.concatenate(([True],np.any(a[ind[1:]]!=a[ind[:-1]],axis=1)))]100 loops, best of 3: 5.93 ms per loopa = np.random.randint(2, size=(10000, 100))%timeit np.unique(a.view(np.dtype((np.void, a.dtype.itemsize*a.shape[1])))).view(a.dtype).reshape(-1, a.shape[1])10 loops, best of 3: 29.9 ms per loop%timeit ind = np.lexsort(a.T); a[np.concatenate(([True],np.any(a[ind[1:]]!=a[ind[:-1]],axis=1)))]10 loops, best of 3: 116 ms per loop