python - numpy.unique generates a list unique in what regard? -

if input array general objects numpy.unique, result unique based upon what?

i have tried:

import numpy np  class a(object): #probably exists nice mixin :p     def __init__(self, a):         self.a =     def __lt__(self, other):         return self.a < other.a     def __le__(self, other):         return self.a <= other.a     def __eq__(self, other):         return self.a == other.a     def __ge__(self, other):         return self.a >= other.a     def __gt__(self, other):         return self.a > other.a     def __ne__(self, other):         return self.a != other.a     def __repr__(self):         return "a({})".format(self.a)     def __str__(self):        return self.__repr__()  np.unique(map(a, range(3)+range(3))) 

which returns

array([a(0), a(0), a(1), a(1), a(2), a(2)], dtype=object) 

but intentions get:

array([a(0), a(1), a(2)], dtype=object) 

assuming duplicate a(2) typo, think need define __hash__ (see docs):

import numpy np functools import total_ordering  @total_ordering class a(object):     def __init__(self, a):         self.a =     def __lt__(self, other):         return self.a < other.a     def __eq__(self, other):         return self.a == other.a     def __ne__(self, other):         return self.a != other.a     def __hash__(self):         return hash(self.a)     def __repr__(self):         return "a({})".format(self.a)     def __str__(self):        return repr(self) 

produces

>>> map(a, range(3)+range(3)) [a(0), a(1), a(2), a(0), a(1), a(2)] >>> set(map(a, range(3)+range(3))) set([a(0), a(1), a(2)]) >>> np.unique(map(a, range(3)+range(3))) array([a(0), a(1), a(2)], dtype=object) 

where i've used total_ordering reduce proliferation of methods, guessed possible. :^)

[edited after posting correct missing __ne__.]