use tf.shape() on tensorflow placeholder - tensorflow

Question

Let's looke at this simple made up tf operation:
data = np.random.rand(1,2,3)
x = tf.placeholder(tf.float32, shape=[None, None, None], name='x_pl')
out = x
print ('shape:', tf.shape(out))
sess = tf.Session()
sess.run(out, feed_dict={x: data})
and the print is:
shape: Tensor("Shape_13:0", shape=(3,), dtype=int32)
I read that you should use tf.shape() to get the 'dynamic' shape of the tensor, which seems to be what I need, but why the shape is shape=(3,)?
why it is not (1,2,3)? as it should be determined when the session is run?
suppose this is part of a neural network where I need to know the last dimension of x, for example, to pass x into a Dense layer, for which the last dimension of x needed to be known.
how do it do it then?

Answer 1

It is because tf.shape() is an op and you have to run it within a session.
data = np.random.rand(1,2,3)
x = tf.placeholder(tf.float32, shape=[None, None, None], name='x_pl')
out = x
print ('shape:', tf.shape(out))
z = tf.shape(out)
sess = tf.Session()
out_, z_ =sess.run([out,z], feed_dict={x: data})
print(f"shape of out: {z_}")
will return
shape: Tensor("Shape:0", shape=(3,), dtype=int32)
shape of out: [1 2 3]
Even if you look at the example from the docs (https://www.tensorflow.org/api_docs/python/tf/shape):
t = tf.constant([[[1, 1, 1], [2, 2, 2]], [[3, 3, 3], [4, 4, 4]]])
tf.shape(t)
If you run it just like that it will return something like
<tf.Tensor 'Shape_4:0' shape=(3,) dtype=int32>
but if you run it within a session then you will get the expected result
t = tf.constant([[[1, 1, 1], [2, 2, 2]], [[3, 3, 3], [4, 4, 4]]])
print(sess.run(tf.shape(t)))
[2 2 3]