功能API中的Keras Multiply()层

Question

在新的API更改下,如何在Keras中逐层实现层的乘法？在旧的API下,我会尝试这样的事情:

merge([dense_all, dense_att], output_shape=10, mode='mul')

我试过这个(MWE):

from keras.models import Model
from keras.layers import Input, Dense, Multiply

def sample_model():
        model_in = Input(shape=(10,))
        dense_all = Dense(10,)(model_in)
        dense_att = Dense(10, activation='softmax')(model_in)
        att_mull = Multiply([dense_all, dense_att]) #merge([dense_all, dense_att], output_shape=10, mode='mul')
        model_out = Dense(10, activation="sigmoid")(att_mull)
        return 0

if __name__ == '__main__':
        sample_model()

完整跟踪:

Using TensorFlow backend.
Traceback (most recent call last):
  File "testJan17.py", line 13, in <module>
    sample_model()
  File "testJan17.py", line 8, in sample_model
    att_mull = Multiply([dense_all, dense_att]) #merge([dense_all, dense_att], output_shape=10, mode='mul')
TypeError: __init__() takes exactly 1 argument (2 given)

编辑:

我尝试实现tensorflow的元素乘法函数.当然,结果不是Layer()实例,所以它不起作用.这是后人的尝试:

def new_multiply(inputs): #assume two only - bad practice, but for illustration...
        return tf.multiply(inputs[0], inputs[1])


def sample_model():
        model_in = Input(shape=(10,))
        dense_all = Dense(10,)(model_in)
        dense_att = Dense(10, activation='softmax')(model_in) #which interactions are important?
        new_mult = new_multiply([dense_all, dense_att])
        model_out = Dense(10, activation="sigmoid")(new_mult)
        model = Model(inputs=model_in, outputs=model_out)
        model.compile(optimizer='adam', loss='categorical_crossentropy', metrics=['accuracy'])
        return model

Answer 1

与keras> 2.0:

from keras.layers import multiply
output = multiply([dense_all, dense_att])

Answer 2

在功能性API下，您只需使用multiply功能，注意小写的“ m”。如您所见，Multiply类是一层，旨在与顺序API一起使用。

有关更多信息，请参见https://keras.io/layers/merge/#multiply_1

Answer 3

您需要在前面再添加一个开/关括号。

from keras.layers import Multiply
att_mull = Multiply()([dense_all, dense_att])