在自定义keras层的调用函数中传递附加参数 [英] Pass additional parameter in call function of custom keras layer
问题描述
我创建了一个定制的keras层,目的是在推理过程中手动更改上一层的激活.以下是简单地将激活次数与数字相乘的基本层.
I created a custom keras layer with the purpose of manually changing activations of previous layer during inference. Following is basic layer that simply multiplies the activations with a number.
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)
当我将其与IRIS数据集一起使用时,效果很好
It works fine when I use it like this with IRIS dataset
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()
但是现在我想将'n'从 init 移至调用函数,以便在训练后可以使用不同的n值来评估模型.这个想法是在n处有一个占位符,可以在调用评估函数之前用一些值来初始化它.我不确定如何实现这一目标.正确的方法是什么?谢谢
However now I want to move 'n' from init to the call function so I can apply different values of n after training to evaluate model. The idea is to have a placeholder inplace of n which can be initialzed with some value before calling the evaluate function on it. I am not sure how to achieve this. What would the correct approach for this? Thanks
推荐答案
You should work the same way the Concatenate layer does. (Search for class Concatenate(_Merge):
in that link).
这些接受多个输入的层依赖于在列表中传递的输入(和输入形状).
These layers taking multiple inputs rely on the inputs (and the input shapes) being passed in a list.
请参见 build
, call
和 comput_output_shape
中的验证部分:
See the verification part in build
, call
and comput_output_shape
:
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,您将其称为列表传递给该层.但是为此,我强烈建议您远离 Sequential
模型.它们纯粹是局限性.
Then you call this layer passing a list to it. But for that, I strongly recommend you move away from Sequential
models. They're pure limitation.
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,并使用 Input(tensor = ...)
,该模型将不会像通常那样要求将2个输入传递给 fit
和 predict
方法.
Using the old answer, with Input(tensor=...)
, the model will not demand, as usually would happen, that you pass 2 inputs to the fit
and predict
methods.
但是使用新选项,并使用 Input(shape = ...)
,它将需要两个输入,因此:
But using the new option, with Input(shape=...)
it will demand two inputs, so:
nArray = np.full((X_train.shape[0],1),n)
model.fit([X_train,nArray],Y_train,....)
不幸的是,我不能使它仅与具有一个元素的 n
一起使用.它必须具有与之完全相同的样本数(这是keras的限制).
Unfortunately, I coulnd't make it work with n
having only one element. It must have exactly the same number of samples as (this is a keras limitation).
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