如何在没有__hash__的情况下删除对象列表中的重复项

问：

我有一个自定义对象列表，我想从中删除重复项。通常，您可以通过定义对象 和 然后获取对象列表来执行此操作。我已经定义了，但我无法想出一种很好的实现方法，以便它为相等的对象返回相同的值。__eq____hash__set__eq____hash__

更具体地说，我有一个派生自 ete3 工具包中的类的类。我定义了两个对象，如果它们的 Robinson-Foulds 距离为零，则它们相等。Tree

from ete3 import Tree

class MyTree(Tree):

    def __init__(self, *args, **kwargs):
        super(MyTree, self).__init__(*args, **kwargs)

    def __eq__(self, other):
        rf = self.robinson_foulds(other, unrooted_trees=True)
        return not bool(rf[0])

newicks = ['((D, C), (A, B),(E));',
           '((D, B), (A, C),(E));',
           '((D, A), (B, C),(E));',
           '((C, D), (A, B),(E));',
           '((C, B), (A, D),(E));',
           '((C, A), (B, D),(E));',
           '((B, D), (A, C),(E));',
           '((B, C), (A, D),(E));',
           '((B, A), (C, D),(E));',
           '((A, D), (B, C),(E));',
           '((A, C), (B, D),(E));',
           '((A, B), (C, D),(E));']

trees = [MyTree(newick) for newick in newicks]

print len(trees)       # 12
print len(set(trees))  # also 12, not what I want!

两者都返回 12，但这不是我想要的，因为几个对象彼此相等：print len(trees)print len(set(trees))

from itertools import product
for t1, t2 in product(newicks, repeat=2):
    if t1 != t2:
        mt1 = MyTree(t1)
        mt2 = MyTree(t2)
        if mt1 == mt2:
            print t1, '==', t2

返回：

((D, C), (A, B),(E)); == ((C, D), (A, B),(E));
((D, C), (A, B),(E)); == ((B, A), (C, D),(E));
((D, C), (A, B),(E)); == ((A, B), (C, D),(E));
((D, B), (A, C),(E)); == ((C, A), (B, D),(E));
((D, B), (A, C),(E)); == ((B, D), (A, C),(E));
((D, B), (A, C),(E)); == ((A, C), (B, D),(E));
((D, A), (B, C),(E)); == ((C, B), (A, D),(E));
((D, A), (B, C),(E)); == ((B, C), (A, D),(E));
((D, A), (B, C),(E)); == ((A, D), (B, C),(E));
((C, D), (A, B),(E)); == ((D, C), (A, B),(E));
((C, D), (A, B),(E)); == ((B, A), (C, D),(E));
((C, D), (A, B),(E)); == ((A, B), (C, D),(E));
((C, B), (A, D),(E)); == ((D, A), (B, C),(E));
((C, B), (A, D),(E)); == ((B, C), (A, D),(E));
((C, B), (A, D),(E)); == ((A, D), (B, C),(E));
((C, A), (B, D),(E)); == ((D, B), (A, C),(E));
((C, A), (B, D),(E)); == ((B, D), (A, C),(E));
((C, A), (B, D),(E)); == ((A, C), (B, D),(E));
((B, D), (A, C),(E)); == ((D, B), (A, C),(E));
((B, D), (A, C),(E)); == ((C, A), (B, D),(E));
((B, D), (A, C),(E)); == ((A, C), (B, D),(E));
((B, C), (A, D),(E)); == ((D, A), (B, C),(E));
((B, C), (A, D),(E)); == ((C, B), (A, D),(E));
((B, C), (A, D),(E)); == ((A, D), (B, C),(E));
((B, A), (C, D),(E)); == ((D, C), (A, B),(E));
((B, A), (C, D),(E)); == ((C, D), (A, B),(E));
((B, A), (C, D),(E)); == ((A, B), (C, D),(E));
((A, D), (B, C),(E)); == ((D, A), (B, C),(E));
((A, D), (B, C),(E)); == ((C, B), (A, D),(E));
((A, D), (B, C),(E)); == ((B, C), (A, D),(E));
((A, C), (B, D),(E)); == ((D, B), (A, C),(E));
((A, C), (B, D),(E)); == ((C, A), (B, D),(E));
((A, C), (B, D),(E)); == ((B, D), (A, C),(E));
((A, B), (C, D),(E)); == ((D, C), (A, B),(E));
((A, B), (C, D),(E)); == ((C, D), (A, B),(E));
((A, B), (C, D),(E)); == ((B, A), (C, D),(E));

所以我的问题是：

• 对于我的情况来说，什么是一个好的实施方式？__hash__set(trees)
• 或者如何在没有定义的情况下从列表中删除相等的对象？__hash__