问题
I built a simple custom layer in Keras and was surprised to find that the parameters were not set to trainable by default. I can get it to work by explicitly setting the trainable attribute. I can't explain why this is by looking at documentation or code. Is this how it is supposed to be or I am doing something wrong which is making the parameters non-trainable by default? Code:
import tensorflow as tf
class MyDense(tf.keras.layers.Layer):
def __init__(self, **kwargs):
super(MyDense, self).__init__(kwargs)
self.dense = tf.keras.layers.Dense(2, tf.keras.activations.relu)
def call(self, inputs, training=None):
return self.dense(inputs)
inputs = tf.keras.Input(shape=10)
outputs = MyDense()(inputs)
model = tf.keras.Model(inputs=inputs, outputs=outputs, name='test')
model.compile(loss=tf.keras.losses.MeanSquaredError())
model.summary()
Output:
Model: "test"
_________________________________________________________________
Layer (type) Output Shape Param #
=================================================================
input_1 (InputLayer) [(None, 10)] 0
_________________________________________________________________
my_dense (MyDense) (None, 2) 22
=================================================================
Total params: 22
Trainable params: 0
Non-trainable params: 22
_________________________________________________________________
If I change the custom layer creation like this:
outputs = MyDense(trainable=True)(inputs)
the output is what I expect (all parameters are trainable):
=================================================================
Total params: 22
Trainable params: 22
Non-trainable params: 0
_________________________________________________________________
then it works as expected and makes all the parameters trainable. I don't understand why that is needed though.
回答1:
No doubt, that's an interesting quirk.
When making a custom layer, a tf.Variable will be automatically included in the list of trainable_variable. You didn't use tf.Variable, but a tf.keras.layers.Dense object instead, which will not be treated as a tf.Variable, and not set trainable=True by default. However, the Dense object you used will be set to trainable. See:
MyDense().dense.trainable
True
If you used tf.Variable (as it should), it will be trainable by default.
import tensorflow as tf
class MyDense(tf.keras.layers.Layer):
def __init__(self, units=2, input_dim=10):
super(MyDense, self).__init__()
w_init = tf.random_normal_initializer()
self.w = tf.Variable(
initial_value=w_init(shape=(input_dim, units), dtype="float32"),
trainable=True,
)
b_init = tf.zeros_initializer()
self.b = tf.Variable(
initial_value=b_init(shape=(units,), dtype="float32"), trainable=True
)
def call(self, inputs, **kwargs):
return tf.matmul(inputs, self.w) + self.b
inputs = tf.keras.Input(shape=10)
outputs = MyDense()(inputs)
model = tf.keras.Model(inputs=inputs, outputs=outputs, name='test')
model.compile(loss=tf.keras.losses.MeanSquaredError())
model.summary()
Model: "test"
_________________________________________________________________
Layer (type) Output Shape Param #
=================================================================
input_11 (InputLayer) [(None, 10)] 0
_________________________________________________________________
my_dense_18 (MyDense) (None, 2) 22
=================================================================
Total params: 22
Trainable params: 22
Non-trainable params: 0
_________________________________________________________________
来源：https://stackoverflow.com/questions/65475110/are-keras-custom-layer-parameters-non-trainable-by-default