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我有一个自定义对象列表，我想从中删除重复项。通常，您可以通过为您的对象定义__eq__和__hash__，然后获取set对象列表中的 来完成此操作。我已经定义__eq__了，但我想不出一个好的实现方法，__hash__以便它为相等的对象返回相同的值。

更具体地说，我有一个派生自ete3 toolkitTree的类的类。如果两个对象的Robinson-Foulds距离为零，我将它们定义为相等。

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

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：

  • 对于我的案例来说，什么是一个好的__hash__实施方案，这样才set(trees)有效？
  • 或者如何在__hash__未定义的情况下从列表中删除相等的对象？
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