'Tensor' object has no attribute '_keras_history'

Question

I wanted to extract and apply independently a Conv2D layer on the columns of my input tensors, but after adding the code:

accelerometer_input = Input(shape=(1400, 3))

for i in range(3):
    out = Lambda(lambda x: x[:,:, i:i+1])(accelerometer_input) # Extracting the ith channel
    out = K.expand_dims(out, axis=1)
    out = Conv2D(64, (30, 1), data_format="channels_first")(out)  
    branch_outputs.append(out)
out_put = K.concatenate(branch_outputs)

it gives me the error in the title. I think it is due to the Lambda layer or the extraction which is not differentiable.

But how can I do without it?

today · Accepted Answer · 2018-10-02 22:59:27Z

3

That's because you are directly applying a backend function (i.e. K.expand_dims()) on a Keras Tensor (i.e. out) and therefore the result would be a Tensor (and not a Keras Tensor). Actually, a Keras Tensor is an augmented version of Tensor and have additional attributes (e.g. _keras_history) which helps Keras to build the model. Now, to resolve the issue, you just need to put the backend function in a Lambda layer to have a Keras Tensor as output:

out = Lambda(lambda x: K.expand_dims(x, axis=1))(out)

The same thing applies to using K.concatenate(). However in this case, there is a specific layer for it in Keras:

from keras.layers import concatenate, Concatenate

# use functional interface
out_put = concatenate(branch_outputs)

# or use layer class
out_put = Concatenate()(branch_outputs)

edited Oct 2, 2018 at 22:59

answered Oct 2, 2018 at 22:22

today

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2 Comments

Francesco Pegoraro Over a year ago

it was: out_put = K.concatenate(branch_outputs), I removed the K.and worked. Your answer helped me understand it, so if you want you can add it to your answer and i'll accept it

today Over a year ago

@FrancescoPegoraro Oh! I did not even see that line. Yeah, it is wrong as well for the same reason. You need to use the Concatenate layer instead. I have updated my answer.

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