您应该使用与“Concatenate”层相同的方法。
这些采用多个输入的层依赖于在列表中传递的输入（和输入形状）。
请参见build、call和comput_output_shape中的验证部分：

def call(self,inputs):
    if not isinstance(inputs, list):
        raise ValueError('This layer should be called on a list of inputs.')
    
    mainInput = inputs[0]
    nInput = inputs[1]

    changed = tf.multiply(mainInput,nInput)
    #I suggest using an equivalent function in K instead of tf here, if you ever want to test theano or another backend later. 
    #if n is a scalar, then just "changed=nInput * mainInput" is ok

    #....the rest of the code....

然后你调用这个层，传递一个列表给它。但是为了这个，我强烈建议你远离Sequential模型。它们是纯粹的限制。

from keras.models import Model

inputTensor = Input((5,)) # the original input (from your input_shape)

#this is just a suggestion, to have n as a manually created var
#but you can figure out your own ways of calculating n later
nInput = Input((1,))
    #old answer: nInput = Input(tensor=K.variable([n]))

#creating the graph
out = Dense(units, input_shape=(5,),activation='relu')(inputTensor)

#your layer here uses the output of the dense layer and the nInput
out = myLayer()([out,nInput])
    #here you will have to handle n with the same number of samples as x. 
    #You can use `inputs[1][0,0]` inside the layer

out = Dense(units,activation='relu')(out)
out = Dense(3,activation='softmax')(out)

#create the model with two inputs and one output:
model = Model([inputTensor,nInput], out)
    #nInput is now a part of the model's inputs

model.compile(optimizer='adam', loss='categorical_crossentropy', metrics=['acc'])

使用Input(tensor=...)的旧答案，模型将不会像通常发生的那样要求您将2个输入传递给fit和predict方法。
但是使用新选项，对于Input(shape=...)，它将需要两个输入，因此：

nArray = np.full((X_train.shape[0],1),n)
model.fit([X_train,nArray],Y_train,....)

不幸的是，我不能让它与只有一个元素的n一起工作。它必须有与完全相同的样本数（这是一个keras限制）。