Tensorflow compute_output_shape() Not Working For Custom Layer

Question

I have created a custom layer (called GraphGather) in Keras, yet the output tensor prints as :

Tensor("graph_gather/Tanh:0", shape=(?, ?), dtype=float32)

For some reason the shape is being returned as (?,?), which is causing the next dense layer to raise the following error:

ValueError: The last dimension of the inputs to Dense should be defined. Found None.

The GraphGather layer code is as follows:

class GraphGather(tf.keras.layers.Layer):

  def __init__(self, batch_size, num_mols_in_batch, activation_fn=None, **kwargs):
    self.batch_size = batch_size
    self.num_mols_in_batch = num_mols_in_batch
    self.activation_fn = activation_fn
    super(GraphGather, self).__init__(**kwargs)

  def build(self, input_shape):
    super(GraphGather, self).build(input_shape)

 def call(self, x, **kwargs):
    # some operations (most of def call omitted)
    out_tensor = result_of_operations() # this line is pseudo code
    if self.activation_fn is not None:
      out_tensor = self.activation_fn(out_tensor)
    out_tensor = out_tensor
    return out_tensor

  def compute_output_shape(self, input_shape):
    return (self.num_mols_in_batch, 2 * input_shape[0][-1])}

I have also tried hardcoding compute_output_shape to be:python def compute_output_shape(self, input_shape): return (64, 150) ``` Yet the output tensor when printed is still

Tensor("graph_gather/Tanh:0", shape=(?, ?), dtype=float32)

which causes the ValueError written above.

System information

Have written custom code
**OS Platform and Distribution*: Linux Ubuntu 16.04
TensorFlow version (use command below): 1.5.0
Python version: 3.5.5

maybe it expects same number of examples in next layer, and problem is in batch size component in shape? — quester
– quester, Commented Jun 9, 2019 at 12:39

Johnny · Accepted Answer · 2018-08-13 14:18:16Z

6

I had the same problem. My workaround was to add the following lines to the call method:

input_shape = tf.shape(x)

and then:

return tf.reshape(out_tensor, self.compute_output_shape(input_shape))

I haven't run into any problems with it yet.

answered Aug 13, 2018 at 14:18

Johnny

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Comments

MarkV · Accepted Answer · 2021-12-16 16:17:12Z

0

If Johnny's answer doesn't work, I found another way to get around this is to follow advice here https://github.com/tensorflow/tensorflow/issues/38296#issuecomment-623698709

which is to call the set_shape method on the output of your layer.

E.g.

l=GraphGather(...)
y=l(x)
y.set_shape( l.compute_output_shape(x.shape) )

This only works if you are using the functional API.

answered Dec 16, 2021 at 16:17

MarkV

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