Keras: Lambda layer function with multiple parameters

Question

I am trying to write a Lambda layer in Keras which calls a function connection, that runs a loop for i in range(0,k) where k is fed in as an input to the function, connection(x,k). Now, when I try to call the function in the Functional API, I tried using:

k = 5
y = Lambda(connection)(x)

Also,

y = Lambda(connection)(x,k)

But neither of those approaches worked. How can I feed in the value of k without assigning it as a global parameter?

Is "k" a constant? Or is it calculated somewhere in the model? Is it an input to the model, as part of the input data? — Daniel Möller
– Daniel Möller, Commented Jul 5, 2017 at 20:42
k updates through the model. The value of k changes for different times I call the Lambda layer. But I found the solution here, in a Keras GitHub Issue. Using y = Lambda(connection, arguments={'k':k})(x) worked! — Prabaha
– Prabaha, Commented Jul 5, 2017 at 22:39

Andrey Nikishaev · Accepted Answer · 2018-07-09 11:07:51Z

22

Just use

y = Lambda(connection)((x,k))

and then var[0], var[1] in connection method

answered Jul 9, 2018 at 11:07

Andrey Nikishaev

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4 Comments

nairouz mrabah Over a year ago

This is the correct solution which should have been accepted.

Arthur Attout Over a year ago

This will not work unless both x and k are Tensor objects (cf :

Layer lambda_4 was called with an input that isn't a symbolic tensor. Received type: <class 'int'>. Full input: [8]. All inputs to the layer should be tensors.

)

Andrey Nikishaev Over a year ago

@ArthurAttout python variables can be passed to Lambda using local context

Daniel Wiczew Over a year ago

I throws error "connection missing 1 required positional argument k" in the newest tensorflow

Prabaha · Accepted Answer · 2017-07-05 22:45:40Z

15

Found the solution to the problem in this GitHub Pull Request. Using

y = Lambda(connection, arguments={'k':k})(x)

worked!

answered Jul 5, 2017 at 22:45

Prabaha

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2 Comments

wrongu Over a year ago

Note: by default model.save_weights() will not recognize k as part of the model.

zwep Over a year ago

Yes, this worked indeed. But for the other readers.. please know that the variable k here is already defined as some constant

Debabrata Pal · Accepted Answer · 2020-03-25 09:30:11Z

4

Tmodel = Sequential()
x = layers.Input(shape=[1,])   # Lambda on single input
out1 = layers.Lambda(lambda x: x ** 2)(x)

y = layers.Input(shape=[1,])   # Lambda on multiple inputs
z = layers.Input(shape=[1,])
def conn(IP):
    return IP[0]+IP[1]
out2 = layers.Lambda(conn)([y,z])

Tmodel = tf.keras.Model(inputs=[x,y,z], outputs=[out1,out2],name='Tmodel')  # Define Model
Tmodel.summary()

# output
O1,O2 = Tmodel([2,15,10])
print(O1)   # tf.Tensor(4, shape=(), dtype=int32)
print(O2)   # tf.Tensor(25, shape=(), dtype=int32)

answered Mar 25, 2020 at 9:30

Debabrata Pal

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1 Comment

Pavel Sem Over a year ago

Please add some explanation for better clarity or reference documentation.

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