Question

我创建了一个自定义keras图层，目的是在推理期间手动更改上一层的激活。以下是简单地将激活与数字相乘的基本层。

import numpy as np
from keras import backend as K
from keras.layers import Layer
import tensorflow as tf

class myLayer(Layer):

    def __init__(self, n=None, **kwargs):
        self.n = n
        super(myLayer, self).__init__(**kwargs)

    def build(self, input_shape):

        self.output_dim = input_shape[1]
        super(myLayer, self).build(input_shape)

    def call(self, inputs):

        changed = tf.multiply(inputs, self.n)

        forTest  = changed
        forTrain = inputs

        return K.in_train_phase(forTrain, forTest)

    def compute_output_shape(self, input_shape):
        return (input_shape[0], self.output_dim)

当我像这样使用IRIS数据集

时，它工作正常

model = Sequential()
model.add(Dense(units, input_shape=(5,)))
model.add(Activation('relu'))
model.add(myLayer(n=3))
model.add(Dense(units))
model.add(Activation('relu'))
model.add(Dense(3))
model.add(Activation('softmax'))
model.compile(optimizer='adam', loss='categorical_crossentropy', metrics=['acc'])
model.summary()

但是现在我想移动＆＃39; n＆＃39;从 init 到调用函数，所以我可以在训练后应用不同的n值来评估模型。我们的想法是在n中放置占位符，在调用其上的evaluate函数之前可以用一些值初始化它。我不知道如何实现这一目标。对此有什么正确的解决方法？感谢

Answer 1

您的工作方式应与Concatenate图层的工作方式相同。（在该链接中搜索class Concatenate(_Merge):）。

这些采用多个输入的图层依赖于在列表中传递的输入（和输入形状）。

请参阅build，call和comput_output_shape中的验证部分：

def call(self,inputs):
    if not isinstance(inputs, list):
        raise ValueError('This layer should be called on a list of inputs.')

    mainInput = inputs[0]
    nInput = inputs[1]

    changed = tf.multiply(mainInput,nInput)
    #I suggest using an equivalent function in K instead of tf here, if you ever want to test theano or another backend later. 
    #if n is a scalar, then just "changed=nInput * mainInput" is ok

    #....the rest of the code....

然后你调用这个图层传递一个列表。但为此，我强烈建议您远离Sequential模型。他们是纯粹的限制。

from keras.models import Model

inputTensor = Input((5,)) # the original input (from your input_shape)

#this is just a suggestion, to have n as a manually created var
#but you can figure out your own ways of calculating n later
nInput = Input((1,))
    #old answer: nInput = Input(tensor=K.variable([n]))

#creating the graph
out = Dense(units, input_shape=(5,),activation='relu')(inputTensor)

#your layer here uses the output of the dense layer and the nInput
out = myLayer()([out,nInput])
    #here you will have to handle n with the same number of samples as x. 
    #You can use `inputs[1][0,0]` inside the layer

out = Dense(units,activation='relu')(out)
out = Dense(3,activation='softmax')(out)

#create the model with two inputs and one output:
model = Model([inputTensor,nInput], out)
    #nInput is now a part of the model's inputs

model.compile(optimizer='adam', loss='categorical_crossentropy', metrics=['acc'])

使用旧答案Input(tensor=...)，模型不会像通常那样要求您将2个输入传递给fit和predict方法。

但是使用新选项Input(shape=...)，它将需要两个输入，所以：

nArray = np.full((X_train.shape[0],1),n)
model.fit([X_train,nArray],Y_train,....)

不幸的是，我不能让n只有一个元素。它必须具有完全相同数量的样本（这是keras限制）。

在自定义keras层的调用函数中传递其他参数

1 个答案: