Tensorflow: access to see the layer activation (fine-tuning),-CodePudding

I use fine-tuning. How can I see and access the activations of all layers that are inside of the convolutional base?

conv_base = VGG16(weights='imagenet',
              include_top=False,
              input_shape=(inp_img_h, inp_img_w, 3))

def create_functional_model():
   inp = Input(shape=(inp_img_h, inp_img_w, 3))
   model = conv_base(inp)
   model = Flatten()(model)
   model = Dense(256, activation='relu')(model)
   outp = Dense(1, activation='sigmoid')(model)
   return Model(inputs=inp, outputs=outp)

model = create_functional_model()
model.summary()

The model summary is

Layer (type)                 Output Shape              Param #   
=================================================================
vgg16 (Functional)           (None, 7, 7, 512)         14714688  
_________________________________________________________________
flatten_2 (Flatten)          (None, 25088)             0         
_________________________________________________________________
dense_4 (Dense)              (None, 256)               6422784   
_________________________________________________________________
dense_5 (Dense)              (None, 1)                 257       
=================================================================
Total params: 21,137,729
Trainable params: 21,137,729
Non-trainable params: 0
_________________________________________________________________

Thus, the levels inside the conv_base are not accessible.

CodePudding user response：

As @Frightera said in comments, you can access the base model summary by:

model.layers[0].summary()

And if you want to access activation functions of its layers you can try this:

print(model.layers[0].layers[index_of_layer].activation)
#or
print(model.layers[0].get_layer("name_of_layer").activation)