Question

构建模型CNN时，输入尺寸为（none，100,100,1），输出尺寸为（400 * 1），但是当我运行模型时，会发生一些错误，这是我的模型：

visible_image1= Input(shape=(100,100,1))
conv_1=Conv2D(filters = 64, kernel_size = (5,5),padding = 'Same',  
             )(visible_image1)
BatchNor_1=BatchNormalization()(conv_1)
relu_1=LeakyReLU(0.2)(BatchNor_1)
pool_1=(MaxPool2D(pool_size=(3,3), strides=(3,3)))(relu_1)
conv_2=Conv2D(filters = 128, kernel_size = (5,5),padding = 'Same', 
             )(pool_1)
BatchNor_2=BatchNormalization()(conv_2)
relu_2=LeakyReLU(0.2)(BatchNor_2)
conv_3=Conv2D(filters = 128, kernel_size = (5,5),padding = 'Same', 
             )(relu_2)
BatchNor_3=BatchNormalization()(conv_3) 
relu_3=LeakyReLU(0.2)(BatchNor_3)  

conv_4=Conv2D(filters = 256, kernel_size = (5,5),padding = 'Same', 
             )(relu_3) 
BatchNor_4=BatchNormalization()(conv_4)  

conv_5=Conv2D(filters = 256, kernel_size = (5,5),padding = 'Same', 
             )( BatchNor_3) 
BatchNor_5=BatchNormalization()(conv_5) 
add_1=Add()([BatchNor_4,BatchNor_5])
relu_4=LeakyReLU(0.2)(add_1)  
conv_6=Conv2D(filters = 128, kernel_size = (5,5),padding = 'Same', 
             )(relu_4)
BatchNor_6=BatchNormalization()(conv_6)  
relu_5=LeakyReLU(0.2)(BatchNor_6) 
conv_7=Conv2D(filters = 128, kernel_size = (5,5),padding = 'Same', 
             )(relu_5)  
BatchNor_7=BatchNormalization()(conv_7) 
relu_6=LeakyReLU(0.2)(BatchNor_7)
conv_8=Conv2D(filters = 256, kernel_size = (5,5),padding = 'Same', 
             )(relu_6) 
BatchNor_8=BatchNormalization()(conv_8)  
add_2=Add()([BatchNor_8, relu_4])
relu_7=LeakyReLU(0.2)(add_2)
conv_9=Conv2D(filters = 128, kernel_size = (5,5),padding = 'Same', 
             )(relu_7)               
BatchNor_9=BatchNormalization()(conv_9) 
relu_8=LeakyReLU(0.2)(BatchNor_9)
conv_10=Conv2D(filters = 128, kernel_size = (5,5),padding = 'Same', 
             )(relu_8) 
BatchNor_10=BatchNormalization()(conv_10) 
relu_9=LeakyReLU(0.2)(BatchNor_10)
conv_11=Conv2D(filters = 256, kernel_size = (5,5),padding = 'Same', 
             )(relu_9)
BatchNor_11=BatchNormalization()(conv_11) 
add_3=Add()([BatchNor_11, relu_7])
relu_10=LeakyReLU(0.2)(add_3)
time_1=TimeDistributed(Dense(256))(relu_10)   #
gru_1=GRU(256, return_sequences=True)(time_1)
flatten_1 = Flatten()(gru_1)  
fc_1=Dense(3000,activation = "relu")(flatten_1)
fc_2=Dense(1000,activation = "relu")(fc_1)
fc_3=Dense(401,activation = "softmax")(fc_2)

错误：

Input 0 is incompatible with layer gru_3: expected ndim=3, found ndim=4

据我所知，relu_10输出的dim是（none 33 33 256），并且在时间分布后，该亮度应该是3D，因为gru层应该有3D输入，我的问题是我应该如何制作尺寸，因为时间分布后是3D层？

时间分布的功能是什么？

Answer 1

我只想重复文章“ enter link description here”的结果，但是我无法解决“时间分布”的问题，我想知道本文中还使用了什么层？ GRU仅用于查找输入数据之间的关系。我真的想知道mu代码有什么问题，因为我对Kears有了新的认识，我已经编码了两个多星期了。

如何在喀拉拉邦使用GRU之前的时间分布？

1 个答案: