在Keras嵌套模型的训练过程中显示哪些损失？

我有一个由3个其他Keras模型（嵌套模型）组成的Keras模型。我的问题是关于Keras培训日志中显示的损失值的含义

以下是我的全球模型摘要：

__________________________________________________________________________________________________
Layer (type)                    Output Shape         Param #     Connected to                     
==================================================================================================
input_16 (InputLayer)           (None, 256, 256, 3)  0                                            
__________________________________________________________________________________________________
model_1 (Model)                 (None, 16, 16, 128)  690368      input_16[0][0]                   
__________________________________________________________________________________________________
model_4 (Model)                 [(None, 17, 4), (None, 17, 4), (None, 16, 16, 128)] 5103826     input_16[0][0]                   
__________________________________________________________________________________________________
concatenate_8 (Concatenate)     (None, 16, 16, 256)  0           model_1[1][0]                    
                                                                 model_4[1][2]                    
__________________________________________________________________________________________________
decoder (Model)                 (None, 256, 256, 3)  582843      concatenate_8[0][0]              
==================================================================================================

这些嵌套模型是2个编码器（

model_1

和

model_4

）和1个解码器（

decoder

）

我还有3个损耗：2个损耗直接应用于

model_4

输出中的2个，一个损耗应用于解码器的输出

当我训练整个模型时，我只看到

model\u 4

一个损失，这被称为

model\u 4\u损失

：

Epoch 34/60
13548/19512 [===================>..........] - ETA: 34:57 - loss: 0.6764 - decoder_loss: 0.0944 - model_4_loss: 0.2797

但是，当我尝试单独训练

model_4

时，我在训练日志中清楚地看到了2个损失（这里

concatenate_xxx

损失对应于

model_4

前2个输出）：

关于这一点，我有几个问题：

• 在培训完整型号时，我是否应该看到3个损失，而不是2个（2个用于

  型号4

  ，1个用于

  解码器

• model_4_loss

  代表什么？来自

  model_4

  的两个损失的平均值？总和？两个损失中只有一个
• 如何使培训日志清楚地显示

  model_4

  的两个损失，而不是一些聚合值

为了提供更多的上下文，以下是我如何构建整个模型的摘要：

encoder1 = build_encoder1()   # returns an object of type `Model` with a single (None, 16, 16, 128) output
encoder2 = build_encoder2()   # returns an object of type `Model` with a list of 3 tensors as output
decoder = build_decoder()     # returns a `Model` with a single (None, 256, 256, 3) output

inp = Input(shape=input_shape)      # input_shape is (None, 256, 256, 3)
z_1 = encoder1(inp)                 # (None, 16, 16, 128)
out1, out2, z_2 = encoder2(inp)     # [(None, 17, 4), (None, 17, 4), (None, 16, 16, 128)]

concat = concatenate[z_1, z_2]      # (None, 16, 16, 256)
out3 = decoder(concat)              # (None, 256, 256, 3)

outputs = [out3, out1, out2]
losses = [loss1(), loss2(), loss2()]     # loss1 is a custom loss function managing the (None, 256, 256, 3) output and loss2 is another managing the (None, 17, 4) outputs
model = Model(inputs=inp, outputs=outputs)
model.compile(loss=losses, optimizer=RMSprop(lr=start_lr))

多谢各位

encoder1 = build_encoder1()   # returns an object of type `Model` with a single (None, 16, 16, 128) output
encoder2 = build_encoder2()   # returns an object of type `Model` with a list of 3 tensors as output
decoder = build_decoder()     # returns a `Model` with a single (None, 256, 256, 3) output

inp = Input(shape=input_shape)      # input_shape is (None, 256, 256, 3)
z_1 = encoder1(inp)                 # (None, 16, 16, 128)
out1, out2, z_2 = encoder2(inp)     # [(None, 17, 4), (None, 17, 4), (None, 16, 16, 128)]

concat = concatenate[z_1, z_2]      # (None, 16, 16, 256)
out3 = decoder(concat)              # (None, 256, 256, 3)

outputs = [out3, out1, out2]
losses = [loss1(), loss2(), loss2()]     # loss1 is a custom loss function managing the (None, 256, 256, 3) output and loss2 is another managing the (None, 17, 4) outputs
model = Model(inputs=inp, outputs=outputs)
model.compile(loss=losses, optimizer=RMSprop(lr=start_lr))