CJR
2020-12-02 05:09:07
If your vector is highly sparse (like 16/30000) I'd probably just skip fiddling with sparse xor entirely.
from scipy import sparse
import numpy as np
import numpy.testing as npt
matrix_1 = sparse.random(10000, 100, density=0.1, format='csc')
matrix_1.data = np.ones(matrix_1.data.shape, dtype=bool)
matrix_2 = sparse.random(1, 100, density=0.1, format='csc', dtype=bool)
vec = matrix_2.A.flatten()
# Pull out the part of the sparse matrix that matches the vector and sum it after xor
matrix_xor = (matrix_1[:, vec].A ^ np.ones(vec.sum(), dtype=bool)[np.newaxis, :]).sum(axis=1)
# Sum the part that doesnt match the vector and add it
l1distances = matrix_1[:, ~vec].sum(axis=1).A.flatten() + matrix_xor
# Double check that I can do basic math
npt.assert_array_equal(l1distances, (matrix_1.A ^ vec[np.newaxis, :]).sum(axis=1))
Thanks a lot it works really well and is 50 times faster than the ^ operation!