numpy.unique 生成的列表在哪些方面是唯一的？

如果您输入一个包含一般对象的数组numpy.unique，结果将是唯一的，基于什么？

我努力了：

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)))

返回

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

但我的意图是得到：

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

假设重复A(2)是一个错字，我认为你只需要定义__hash__（参见docs http://docs.python.org/2/reference/datamodel.html#object.__hash__):

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)

我用过的地方total_ordering http://docs.python.org/2/library/functools.html#functools.total_ordering减少方法的扩散，正如您猜测的那样是可能的。 :^)

[发布后编辑以更正缺失的内容__ne__.]