在Python中将functools.partial设置为实例方法

在Python中将functools.partial设置为实例方法,python,class,methods,closures,Python,Class,Methods,Closures,我正在使用functools.partial创建闭包,并使用setattr从类实例调用make。这里的想法是在运行时创建一组方法 #!/usr/bin/python from functools import partial class MyClass(object): def __init__(self, val): self.val = val @classmethod def generateMethods(self): def

我正在使用
functools.partial
创建闭包,并使用
setattr
从类实例调用make。这里的想法是在运行时创建一组方法

#!/usr/bin/python
from functools import partial

class MyClass(object):

    def __init__(self, val):
        self.val = val

    @classmethod
    def generateMethods(self):
        def dummy(conf1, self):
            print "conf1:", conf1
            print "self.val:", self.val
            print

        for s in ('dynamic_1', 'dynamic_2'):
            closed = partial(dummy, s)
            setattr(self, "test_{0}".format(s), closed)
在我看来,
partial
会将
s
的当前值绑定到
dummy
的第一个参数,这将释放
self
以在从实例调用时传递

这不是我所期望的那样

if __name__ == '__main__':
    # Dynamically create some methods
    MyClass.generateMethods()

    # Create an instance
    x = MyClass('FOO')

    # The dynamically created methods aren't callable from the instance :(
    #x.test_dynamic_1()
    # TypeError: dummy() takes exactly 2 arguments (1 given)

    # .. but these work just fine
    MyClass.test_dynamic_1(x)
    MyClass.test_dynamic_2(x)

是否可以动态创建闭包方法,但可以从类的实例调用这些方法?

问题是,当您使用实例调用它们时,它们实际上不是绑定方法,即它们不知道该实例。绑定方法在调用时自动将
self
插入到基础函数的参数中,它存储在绑定方法的
\uuuuuuuuuuuuuuuuuuuuuuuu
属性中

因此,重写并查看正在获取的对象是否是
partial
类型的实例,如果是,则使用将其转换为绑定方法

代码:

#!/usr/bin/python
from functools import partial
import types


class MyClass(object):

    def __init__(self, val):
        self.val = val

    @classmethod
    def generateMethods(self):
        def dummy(conf1, self): 
            print "conf1:", conf1
            print "self.val:", self.val
            print

        for s in ('dynamic_1', 'dynamic_2'):
            closed = partial(dummy, s)
            setattr(self, "test_{0}".format(s), closed)

    def __getattribute__(self, attr):
        # Here we do have access to the much need instance(self)
        obj = object.__getattribute__(self, attr)
        if isinstance(obj, partial):    
            return types.MethodType(obj, self, type(self))
        else:
            return obj


if __name__ == '__main__':
    MyClass.generateMethods()

    x = MyClass('FOO')

    x.test_dynamic_1()
    x.test_dynamic_2()
我认为新的是为了这个确切的用例

直接从文档中:

>>> class Cell(object):
...     def __init__(self):
...         self._alive = False
...     @property
...     def alive(self):
...         return self._alive
...     def set_state(self, state):
...         self._alive = bool(state)
...     set_alive = partialmethod(set_state, True)
...     set_dead = partialmethod(set_state, False)
...
>>> c = Cell()
>>> c.alive
False
>>> c.set_alive()
>>> c.alive
True

可能重复@alfasin,该问题询问如何向特定实例添加方法;我的问题是关于向类中添加方法的问题,这些方法将由所有instancesTrue继承,但公认的答案显示了如何完成这两项工作!我想没有办法立即将闭包添加为
MethodType
,而不是在调用时动态转换它?