测试实例是否来自同一个类

Question

我有一个主要的“抽象”类宠物，和两个“真实”类狗和猫。当我有两个宠物的例子时，我想知道它们是否是“同一种宠物”，而不考虑它们是哪种宠物。

我试过这个

#!/usr/bin/python

class Pet:
    name = "pet"

    def __eq__(self, other):
        return type(self) == type(other)
#        return type(self).__name__ == type(other).__name__
#        return self.__class__ == other.__class__
#        return self.__class__.__name__ == other.__class__.__name__
#        return self.__class__ is other.__class__
#        return self.__class__.__name__ is other.__class__.__name__
#        return isinstance(other, type(self).__name__)
#        return isinstance(other, type(self))
#        return isinstance(self, type(other).__name__)
#        return isinstance(self, type(other))
#        return type(self) is type(other)
#        return type(self).__name__ is type(other).__name__

class Dog:
    name = "dog"

class Cat:
    name = "cat"

dog1 = Dog()
dog2 = Dog()
cat1 = Cat()
cat2 = Cat()

print("dog1 == dog2: {0}".format(dog1 == dog2))
print("cat1 == cat2: {0}".format(cat1 == cat2))
print("cat1 == dog1: {0}".format(cat1 == dog1))

我尝试了所有评论的返回，它总是给出这样的结果：

$ ./test.py 
dog1 == dog2: False
cat1 == cat2: False
cat1 == dog1: False

有没有办法在主类中进行这样的测试以避免代码重复？

编辑：
多人提到它忘记了子类化部分
这可以按预期工作

#!/usr/bin/python

class Pet:
    name = "pet"

    def __eq__(self, other):
        return self.__class__ == other.__class__

class Dog(Pet):
    name = "dog"

class Cat(Pet):
    name = "cat"

Answer 1

您可以使用类__名称用于此：

dog1__name__ == dog2.__name__

使现代化

正如@Tbaki所说，您也可以使用type（）：

type(dog1) == type(dog2)

另一个更新

type(dog1) is type(dog2) # notice that I use 'is'

Answer 2

一种方法是使用上面代码中定义的name参数。另外，Dog

#!/usr/bin/python

class Pet:
    name = "pet"

    def __eq__(self, other):
        return self.name == other.name


class Dog(Pet):
    name = "dog"

class Cat(Pet):
    name = "cat"

dog1 = Dog()
dog2 = Dog()
cat1 = Cat()
cat2 = Cat()

print("dog1 == dog2: {0}".format(dog1 == dog2))
print("cat1 == cat2: {0}".format(cat1 == cat2))
print("cat1 == dog1: {0}".format(cat1 == dog1))