Keras Functional API 层不兼容问题答案

【问题标题】：Keras Functional API imcompatible layer problemKeras Functional API 层不兼容问题
【发布时间】：2021-12-15 18:53:56
【问题描述】：

我试图用代码制作 DDPG_critic 的神经网络层，

def get_critic():
    #num_states = 8 ; num_actions = 2
    state_input = Input(shape=(num_states,),name='critic_state_input_layer')
    state_out = Dense(32, activation="relu",name='critic_state_output_layer')(state_input)

    action_input = Input(shape=(num_actions,),name='critic_action_input_layer')
    action_out = Dense(32,activation="relu",name='critic_action_output_layer')(action_input)

    concat = layers.Concatenate(axis=-1)([state_out, action_out])

    out3 = Dense(256, activation="relu",name='critic_out3_layer')(concat)
    out4 = Dense(256, activation="relu",name='critic_out4_layer')(out3)
    outputs = Dense(1,name='critic_output_layer')(out4)

    model = Model([state_input, action_input], outputs,name='critic_model')

我有问题

ValueError: Exception encountered when calling layer "critic_model" (type Functional).

Input 0 of layer "critic_action_output_layer" is incompatible with the layer: expected axis -1of input shape to have value 2, but received input with shape (64, 1)

如果您指出问题以及如何解决，将不胜感激！

【问题讨论】：

标签： python tensorflow keras reinforcement-learning valueerror

【解决方案1】：

模型架构没有问题。检查您的输入数据形状

import tensorflow as tf
state_input = tf.keras.Input(shape=(8,),name='critic_state_input_layer')
state_out = tf.keras.layers.Dense(32, activation="relu",name='critic_state_output_layer')(state_input)
state_out.shape

输出

TensorShape([None, 32])

第二层

action_input = tf.keras.Input(shape=(2,),name='critic_action_input_layer')
action_out = tf.keras.layers.Dense(32,activation="relu",name='critic_action_output_layer')(action_input)
action_out.shape

输出

TensorShape([None, 32])

【讨论】：