首先:使用<代码>True和(False(boolean property)而不是 页: 1 False (string nature).

Generally, you can make everything comparable in Python. You just have to define specific methods (like __eq__, __lt__, etc.) for your class.

因此,我要说的是,我要比较A类的情况,而比较应当只是对<条码>/代码>成员的个案敏感地比较:

class A(object):
    def __init__(self, s=  ):
        self.s = s

    def __eq__(self, other):
        return self.s.lower() == other.s.lower()

a = A( aaaa )
b = A( AAAA )
print a == b # prints True
b = A( bbbb )
print a == b # prints False

Instead of using instances of Category (e.g. Category( Analyze , True , False , False , FINAL )) as the keys in your dictionary, it sounds like you should be using the associated tuple (e.g. ( Analyze , True , False , False , FINAL )).

If you really do want to use instance of Category as the keys in the dictionary, you ll need to define both the __hash__ and __eq__ methods. For example:

class Category(object):
    def __init__(self,*args):
        self.args=args
    def __hash__(self):
        # Note that this assumes that Category s hash value is immutable
        # i.e. self.args does not change.
        return hash(self.args)
    def __eq__(self,other):
        return self.args == other.args

a=Category( Analyze ,  True ,  False ,  False ,  FINAL )
b=Category( Analyze ,  True ,  False ,  False ,  FINAL )

mydict={a:1}

<代码>a和b是不同的事例,因此它们有不同的<代码>ids,但它们的散射数值相同:

assert id(a) != id(b)
assert hash(a)==hash(b)

b是mydict可接受的关键:

print(mydict[a])
# 1
print(mydict[b])
# 1

PS。 http://docs.python.org/release/3.0.1/whatsnew/3.0.html#ordering-comparisons”rel=“nofollow”

The cmp() function should be treated as gone, and the __cmp__() special method is no longer supported. Use __lt__() for sorting, __eq__() with __hash__(), and other rich comparisons as needed.

由于你显然能够把你的类别事例列入一个字典,你必须拥有已经过时的“条码”_hash_。 现在需要的是<代码>_eq__:

class Category(object):
    # you must have overwritten `__hash__` like this already
    def __hash__(self):
        return hash((self.attr1, self.attr2, ... ))

    # now you just need tis
    def __eq__(self, other):
        return isinstance(other, Category) and 
               (self.attr1, self.attr2, ... ) == (other.attr1, other.attr2, ... )

你们应该做的是,放弃整个类别,使用<条码>,代之以:

Category = collections.namedtuple( Category ,  attr1 attr2 attr3 )

可能有一些捷径,如使用胶卷,但回答贵方问题的一般答案是:执行__eq__(<>>>/code>,供您的班级使用,以便(......)使用,而不是测试身份。