numpy.unique生成一个独特的列表在哪方面？

Question

If you input an array with general objects to numpy.unique , the result will be unique based upon what?

I have tried:

import numpy as np

class A(object): #probably exists a nice mixin for this :P
    def __init__(self, a):
        self.a = 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 my intentions are to get:

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

Answer 1

Assuming the duplicate A(2) is a typo, I think you simply need to define __hash__ (see the docs ):

import numpy as np
from functools import total_ordering

@total_ordering
class A(object):
    def __init__(self, a):
        self.a = 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 to reduce the proliferation of methods, as you guessed was possible. :^)

[Edited after posting to correct missing __ne__ .]