如何使用“ tf.shape（）”获得“静态”形状？

Question

I am just trying to get the static shape of a tensor using command 'tf.shape()'. I noticed that we could use 'x.get_shape' to obtain the static shape, however, in my code, for training, the shape is static, for evaluation, the shape of tensor is dynamic 'None'-so that I am not sure if I also could get static shape by using 'tf.shape()' as well.

The shape of 'deconv4.outputs' is [16,256,256,3], however, when I use 'tf.shape(deconv4.outputs)[0]', the result is: 'Tensor("FRUC_DVF_g/strided_slice_2:0", shape=(), dtype=int32, device=/device:GPU:0)', not '16'.

Could you please help me with this?

Thanks in advance!

I have got the output of the model called 'deconv4.outputs' with the shape of [16,256,256,3]. I want to obtain the first dimension of 'deconv4.outputs' using 'tf.shape()' command.

flow = tf.slice(deconv4.outputs, [0,0,0,0], [tf.shape(deconv4.outputs)[0], 256, 256, 2])

The expected result is that the 'tf.shape(deconv4.outputs)[0]' returns '16'.

Answer 1

tf.shape(), should return 16, when running under a session with appropriate inputs.

Anything which is dynamic, if you specify None at the input, without running it under a session you would get ().

Try your code with eager_execution_enabled if possible.