Question

我正在执行以下Python代码：

class Pet :
    kind = 'dog'    # class variable shared by all instances
    tricks = []

    def __init__(self, name) :
        self.name = name # instance variable unique to each instance
    def changePet(self, newPet) :
        self.kind = newPet
    def addTricks(self, tricks) :
        self.tricks.append(tricks)

pet1 = Pet('mypet')
pet2 = Pet('yourpet')
print 'pet1 kind ::: ', pet1.kind;print 'pet2 kind ::: ', pet2.kind
print 'pet1 name ::: ', pet1.name;print 'pet2 name ::: ', pet2.name

Pet.kind = 'cat'
print 'changed Pet.kind to cat'
print 'pet1 kind ::: ', pet1.kind;print 'pet2 kind ::: ', pet2.kind

#changing pet#1 kind does not change pet#2 kind
pet1.changePet('parrot')
print 'changed pet1.kind to parrot'
print 'pet1 kind ::: ', pet1.kind;print 'pet2 kind ::: ', pet2.kind

pet1.addTricks('imitate')
pet2.addTricks('roll over')
print 'pet1 tricks ::: ', pet1.tricks;print 'pet2 tricks ::: ', pet2.tricks

print Pet.__dict__
print pet1.__dict__
print pet2.__dict__

输出是根据我在互联网上找到的解释。输出如下

pet1 kind :::  dog
pet2 kind :::  dog
pet1 name :::  mypet
pet2 name :::  yourpet
changed Pet.kind to cat
pet1 kind :::  cat
pet2 kind :::  cat
changed pet1.kind to parrot
pet1 kind :::  parrot
pet2 kind :::  cat
pet1 tricks :::  ['imitate', 'roll over']
pet2 tricks :::  ['imitate', 'roll over']
{'__module__': '__main__', 'tricks': ['imitate', 'roll over'], 'kind': 'cat', 'addTricks': <function addTricks at 0xb71fa6bc>, 'changePet': <function changePet at 0xb71fa33c>, '__doc__': None, '__init__': <function __init__ at 0xb71fa144>}
{'kind': 'parrot', 'name': 'mypet'}
{'name': 'yourpet'}

现在我的查询是为什么可变类对象的处理方式与非可变类对象不同

Answer 1

我不确定他们对待的方式不同。您只是对它们执行不同的操作。

对于changePet，您要将之前未定义的self.kind分配给值。现在，当python试图找到pet1.kind时，它会立即找到它。查找pet2.kind时，它找不到它。但是，它确实找到了Pet.kind，因此它会返回。

如果是addTricks，您正试图改变self.tricks。由于这不存在，因此您可以参考Pet.tricks。因此，当您在append()上致电self.tricks时，您可以有效地致电Pet.tricks.append()，而不是self.tricks.append()。

为了澄清事情，请：

def changePet(self, newPet) :
    self.kind = newPet
def addTricks(self, tricks) :
    self.tricks.append(tricks)

实际上等同于：

def changePet(self, newPet) :
    self.kind = newPet
def addTricks(self, tricks) :
    Pet.tricks.append(tricks)

为了证明python不是以不可变的方式处理mutable，我们需要更改你的方法，以便它们执行类似的操作：

def changePet(self, newPet) :
    self.kind = newPet

def addTricks(self, tricks):
    # assign self.tricks to a new value where we previously only mutated it!
    # note: list(self.tricks) returns a copy
    self.tricks = list(self.tricks)
    self.tricks.append(tricks)

现在，当我们再次运行您的代码时，我们得到以下输出：

pet1 kind :::  dog
pet2 kind :::  dog
pet1 name :::  mypet
pet2 name :::  yourpet
changed Pet.kind to cat
pet1 kind :::  cat
pet2 kind :::  cat
changed pet1.kind to parrot
pet1 kind :::  parrot
pet2 kind :::  cat
pet1 tricks :::  ['imitate']
pet2 tricks :::  ['roll over']
{'__module__': '__main__', 'tricks': [], 'kind': 'cat', 'addTricks': <function addTricks at 0x02A33C30>, 'changePet': <function changePet at 0x02A33BF0>, '__doc__': None, '__init__': <function __init__ at 0x02A33BB0>}
{'tricks': ['imitate'], 'kind': 'parrot', 'name': 'mypet'}
{'tricks': ['roll over'], 'name': 'yourpet'}

在python中如何处理可变类属性

1 个答案: