Question

我的基类如下:

class NeuralNetworkBase:
    def __init__(self, numberOfInputs, numberOfHiddenNeurons, numberOfOutputs):
        self.inputLayer = numpy.zeros(shape = (numberOfInputs))
        self.hiddenLayer = numpy.zeros(shape = (numberOfHiddenNeurons))
        self.outputLayer = numpy.zeros(shape = (numberOfOutputs))

        self.hiddenLayerWeights = numpy.zeros(shape = (numberOfInputs, numberOfHiddenNeurons))
        self.outputLayerWeights = numpy.zeros(shape = (numberOfHiddenNeurons, numberOfOutputs))

现在,我有一个精干的班子,其代码如下:

class NeuralNetworkBackPropagation(NeuralNetworkBase):
    def __init__(self, numberOfInputs, numberOfHiddenNeurons, numberOfOutputs):
        self.outputLayerDeltas = numpy.zeros(shape = (numberOfOutputs))
        self.hiddenLayerDeltas = numpy.zeros(shape = (numberOfHiddenNeurons))

但是,当我看一看NeuralNetworkBackPropagation时,我就象两个建筑商一样被叫来。这就是说,我不想凌驾于基类建筑商之上。如果没有,基类构造者在管理衍生类别时是否会 s? 我是否必须在衍生的阶级构造中含蓄地这样做?

Answer 1

Does python call by default the base class constructor s when running the derived class one? Do I have to implicitly do it inside the derived class constructor?

无,是。

这符合“沙尔”处理其他压倒性的方法——如果你想要在继承类别中使用这种功能的话,你必须明确把任何方法从原产类别中删除。

你们的建筑者应当研究这样的情况:

def __init__(self, numberOfInputs, numberOfHiddenNeurons, numberOfOutputs):
    NeuralNetworkBase.__init__(self, numberOfInputers, numberOfHiddenNeurons, numberOfOutputs)
    self.outputLayerDeltas = numpy.zeros(shape = (numberOfOutputs))
    self.hiddenLayerDeltas = numpy.zeros(shape = (numberOfHiddenNeurons))

或者,您可使用<super功能实现同样的目的,但rel=“noreferer” 您在使用时需要小心。

Answer 2

您必须将此列入<条码>_init__(>,即“条码”_init__(<>条/代码>父母类别(NeuralNetworkBase):

NeuralNetworkBase.__init__(self, numberOfInputs, numberOfHiddenNeurons, numberOfOutputs)

家长班的施工者总是被自动称为,除非在子女班子里超婚。如果你在孩子的班子里超婚,想叫父母的班子,那么,正如我前面所显示的那样,你不得不这样做。

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