python - Find unique columns and column membership -

i went through these threads:

• find unique rows in numpy.array
• removing duplicates in each row of numpy array
• pandas: unique dataframe

and discuss several methods computing matrix unique rows , columns.

however, solutions bit convoluted, @ least untrained eye. here example top solution first thread, (correct me if wrong) believe safest , fastest:

np.unique(a.view(np.dtype((np.void, a.dtype.itemsize*a.shape[1])))).view(a.dtype).reshape(-1,  a.shape[1]) 

either way, above solution returns matrix of unique rows. looking along original functionality of np.unique

u, indices = np.unique(a, return_inverse=true) 

which returns, not list of unique entries, membership of each item each unique entry found, how can columns?

here example of looking for:

array([[0, 2, 0, 2, 2, 0, 2, 1, 1, 2],        [0, 1, 0, 1, 1, 1, 2, 2, 2, 2]]) 

we have:

u       = array([0,1,2,3,4]) indices = array([0,1,0,1,1,3,4,4,3]) 

where different values in u represent set of unique columns in original array:

0 -> [0,0] 1 -> [2,1] 2 -> [0,1] 3 -> [2,2] 4 -> [1,2] 

essentially, want np.unique return indexes of unique columns, , indices of they're used? easy enough transposing matrix , using code other question, addition of return_inverse=true.

at = a.t b = np.ascontiguousarray(at).view(np.dtype((np.void, at.dtype.itemsize * at.shape[1]))) _, u, indices = np.unique(b, return_index=true, return_inverse=true) 

with a, gives:

in [35]: u out[35]: array([0, 5, 7, 1, 6])  in [36]: indices out[36]: array([0, 3, 0, 3, 3, 1, 4, 2, 2, 4]) 

it's not entirely clear me want u be, however. if want unique columns, use following instead:

at = a.t b = np.ascontiguousarray(at).view(np.dtype((np.void, at.dtype.itemsize * at.shape[1]))) _, idx, indices = np.unique(b, return_index=true, return_inverse=true) u = a[:,idx] 

this give

in [41]: u out[41]: array([[0, 0, 1, 2, 2],        [0, 1, 2, 1, 2]])  in [42]: indices out[42]: array([0, 3, 0, 3, 3, 1, 4, 2, 2, 4])