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, Jul 05 '17 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](https://github.com/fchollet/keras/pull/1911), in a Keras GitHub Issue. Using `y = Lambda(connection, arguments={'k':k})(x)` worked! — Prabaha, Jul 05 '17 at 22:39

score 21 · Accepted Answer · answered Jul 09 '18 at 11:07

21

Just use

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

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

answered Jul 09 '18 at 11:07

Andrey Nikishaev

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This is the correct solution which should have been accepted. – nairouz mrabah Dec 08 '18 at 19:09
1

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: . Full input: [8]. All inputs to the layer should be tensors.`) – Arthur Attout May 03 '19 at 11:07
@ArthurAttout python variables can be passed to Lambda using local context – Andrey Nikishaev May 04 '19 at 12:09
I throws error "connection missing 1 required positional argument k" in the newest tensorflow – Daniel Wiczew Nov 13 '22 at 17:48

score 15 · Answer 2 · answered Jul 05 '17 at 22:45

15

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

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

worked!

answered Jul 05 '17 at 22:45

Prabaha

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Note: by default `model.save_weights()` will not recognize `k` as part of the model. – wrongu Nov 02 '17 at 19:34
2

Yes, this worked indeed. But for the other readers.. please know that the variable k here is already defined as some constant – zwep Oct 30 '18 at 09:11

score 4 · Answer 3 · answered Mar 25 '20 at 09:30

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)

Please add some explanation for better clarity or reference documentation. — mybrave, Mar 25 '20 at 09:50

Keras: Lambda layer function with multiple parameters

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