How to I factorize a list of tuples?

Question

definition
factorize: Map each unique object into a unique integer. Typically, the range of integers mapped to is from zero to the n - 1 where n is

Answer 1

Approach #1

Convert each tuple to a row of a 2D array, view each of those rows as one scalar using the views concept of NumPy ndarray and finally use np.unique(... return_inverse=True) to factorize -

np.unique(get_row_view(np.array(tups)), return_inverse=1)[1]

get_row_view is taken from here.

Sample run -

In [23]: tups
Out[23]: [(1, 2), ('a', 'b'), (3, 4), ('c', 5), (6, 'd'), ('a', 'b'), (3, 4)]

In [24]: np.unique(get_row_view(np.array(tups)), return_inverse=1)[1]
Out[24]: array([0, 3, 1, 4, 2, 3, 1])

Approach #2

def argsort_unique(idx):
    # Original idea : https://stackoverflow.com/a/41242285/3293881 
    n = idx.size
    sidx = np.empty(n,dtype=int)
    sidx[idx] = np.arange(n)
    return sidx

def unique_return_inverse_tuples(tups):
    a = np.array(tups)
    sidx = np.lexsort(a.T)
    b = a[sidx]
    mask0 = ~((b[1:,0] == b[:-1,0]) & (b[1:,1] == b[:-1,1]))
    ids = np.concatenate(([0], mask0  ))
    np.cumsum(ids, out=ids)
    return ids[argsort_unique(sidx)]

Sample run -

In [69]: tups
Out[69]: [(1, 2), ('a', 'b'), (3, 4), ('c', 5), (6, 'd'), ('a', 'b'), (3, 4)]

In [70]: unique_return_inverse_tuples(tups)
Out[70]: array([0, 3, 1, 2, 4, 3, 1])