将方法作为变量分配给对象

Question

我希望对象在它们的例程中调用某个方法（在变量中分配的特定方法）。我设法用字典做到了，我还设法通过在对象创建后分配变量来做到这一点。哪种方式更好？还有其他更清洁的方法吗？

第一种可行的方法（dict）：

class Foo:
    def __init__(self, name, funcin) -> None:
        self.name = name
        self.func = {
            'bear': self.bear,
            'boar': self.boar,
        }[funcin]
    
    def routine(self):
        print('fluff')
        self.func()

    def bear(self):
        print(f'I, {self.name}, am a bear and I like honey.')
    
    def boar(self):
        print(f'I, {self.name}, am a boar and I like fruits.')

bob = Foo('bob', 'bear')
mark = Foo('mark', 'boar')

bob.routine()
mark.routine()

第二种方法有效（分配另一行）：

class Foo:
    def __init__(self, name) -> None:
        self.name = name
    
    def routine(self):
        print('fluff')
        self.func()

    def bear(self):
        print(f'I, {self.name}, am a bear and I like honey.')
    
    def boar(self):
        print(f'I, {self.name}, am a boar and I like fruits.')
    
Kona = Foo('Kona')
Kona.func = Kona.bear
John = Foo('John')
John.func = John.boar

Kona.routine()
John.routine()

无效的方法：

# NameError: name 'bear' is not defined
class Foo:
    def __init__(self, name, funcin) -> None:
        self.name = name
        self.func = self.funcin
    def bear(self):
        print(f'I, {self.name}, am a bear and I like honey.')
bob = Foo('bob', bear)
bob.func()

# NameError: name 'self' is not defined
class Foo:
    def __init__(self, name, funcin) -> None:
        self.name = name
        self.func = funcin
    def bear(self):
        print(f'I, {self.name}, am a bear and I like honey.')
bob = Foo('bob', self.bear)
bob.func()

# NameError: name 'Kim' is not defined
class Foo:
    def __init__(self, name, funcin) -> None:
        self.name = name
        self.func = funcin
    def bear(self):
        print(f'I, {self.name}, am a bear and I like honey.')
Kim = Foo('Kim', Kim.bear)
Kim.func()   

# TypeError: Foo.bear() missing 1 required positional argument: 'self'
class Foo:
    def __init__(self, name, funcin) -> None:
        self.name = name
        self.func = funcin
    def bear(self):
        print(f'I, {self.name}, am a bear and I like honey.')
bob = Foo('bob', Foo.bear)
bob.func()

Answer 1

class 的方法可以通过 class 的 __dict__ 属性在字典中使用。因此，在您提出的第一个解决方案中，您还可以这样定义 self.func：

self.func = Foo.__dict__[funcin].__get__(self)

所以使用 __dict__[funcin] 我们从 class 对象中检索方法，使用 __get__ 我们得到一个具有正确 self 绑定的函数。

让它更通用，这样你就不用再提了 Foo:

self.func = self.__class__.__dict__[funcin].__get__(self)

Answer 2

一个常见的解决方案是class 继承。从 abstract base classes 到 mixins 有多种选择，但这里是一个简单的开始

class Foo:
    def __init__(self, name) -> None:
        self.name = name
        
    def func(self):
        raise NotImplementedError
    
    def routine(self):
        print('fluff')
        self.func()

class Bear(Foo):
    
    def func(self):
        print(f'I, {self.name}, am a bear and I like honey.')
    
class Boar(Foo):
    
    def func(self):
        print(f'I, {self.name}, am a boar and I like fruits.')

bob = Bear('bob')
mark = Boar('mark')

bob.routine()
mark.routine()

Answer 3

你可以使用 getattr():

class Foo:
    def __init__(self, name, funcin) -> None:
        self.name = name
        self.func = getattr(self,funcin)
    
    def routine(self):
        print('fluff')
        self.func()

    def bear(self):
        print(f'I, {self.name}, am a bear and I like honey.')
    
    def boar(self):
        print(f'I, {self.name}, am a boar and I like fruits.')

如果您不需要执行额外的（常见的）绒毛，您可以简单地分配 self.routine = getattr(self,funcin) 甚至不定义 routine() 函数。

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