python2.7, 新手问题, 代码如下:
class A(object):
def __init__(self):
pass
a = A()
b = A()
为什么a == A()
,a == b
的返回结果都为False
?
感觉很不合逻辑啊? 这是为什么?
收到解答, 那么又有了新的问题, 代码如下:
class A(object):
def __init__(self, *args):
self.args = args
def __eq__(self, another):
if type(self) == type(another) and self.__dict__ == self.__dict__:
return True
return False
a = A(1)
b = A(1)
print a==b
print [a] == [b]
print set([a]) == set([b])
结果为 True, True, False 为什么set
会出现不一样的情况?
a == A()
a == b
实际上是调用__eq__操作符重载来监测对象是否相等,可见
class A(object):
def __init__(self):
pass
def __eq__(self, other):
print "__eq__ called"
return True
a = A()
b = A()
print (a == A())
print(a == b)
监测对象地址用id()函数,比如 id(a) == id(b) 打印a和b是否指向一个地址, 比较对象内容和对象地址是2个意思,见
class A(object):
def __init__(self, _id, _name, _age):
self.mID = _id
self.mName = _name
self.mAge = _age
def __eq__(self, other):
return self.mID == other.mID
a = A(123, "tom", 23)
b = A(123, "jerry", 23)
c = A(234, "jerry", 23)
print a == b
print a == c
print id(a) == id(b)
附加,更新后的代码有几个问题
1) 判断语句应该是if type(self) == type(another) and self.__dict__ == another.__dict__,后面不是self,笔误
2) 你可以在__eq__里加上print "__eq__ called",明显是转换成set之后比较不是用这个__eq__,而是__hash__方法
3) 尽量不要用__dict__吧,个人觉得你显式用args已经很好了
4) 我个人不太喜欢用type来检查类型,更喜欢下面的方式
def __eq__(self, another):
assert isinstance(another, self.__class__)
return self.args == another.args
List 比较用的是 __eq__
Set 比较用的是__hash__
, 因为本质上 set 是个建立在 dict(哈希表)上的结构.
你需要重载__hash__
.
See more http://stackoverflow.com/questions/15326985/how-to-implement-eq-for-set-inclusion-test