在Python中创建一个对象之前,如何使成员函数不确定?

实际上,您可以只传递函数本身,因此您的类将独立于激活函数:

def Signum(x):
    return 1 if x > 0 else 0 

def Sigmoid(x):
    return math.tanh(x) 

def Sin(x):
    return math.sin(x)


class Node():
    '''  
    This is class of one node

    Each node contains an activation function.
    '''  
    def __init__(self, activation=Sigmoid):
        #possible functions
        self.__activate = activation                                       

     def Activation(self, input_value):
         return self.__activate(input_value)

Sigmoid 默认情况下,还要注意,您可以检查激活函数是否是可调用的:

    def __init__(self, activation=Sigmoid):
        #possible functions
        if not callable(actvation):
            raise ValueError("activation should be a callable function")
        self.__activate = activation 

如果您确实需要在kwargs中传递函数,只需检查您需要的密钥是否在其中:

    def __init__(self, **kwargs):
        #possible functions
        activation = None
        if "activation" in kwargs:
            activation = kwargs["activation"]
        if not callable(actvation):
            raise ValueError("activation should be a callable function")
        self.__activate = activation 

class Node:

    def __init__(self, func)
         self.activation = func

那么当你这么做的时候 Node(func).activation() 它将运行 func

我认为这比将函数名作为参数传递要好,因为参数更容易出错