Sometimes the shape of a tensor depends on a value that is computed at runtime. Let’s take the following example, where x is defined as a tf.placeholder() vector with four elements:
x = tf.placeholder(tf.int32, shape=[4])
print x.get_shape()
# ==> '(4,)'
The value of x.get_shape() is the static shape of x, and the (4,) means that it is a vector of length 4. Now let’s apply the tf.unique() op to x
y, _ = tf.unique(x)
print y.get_shape()
# ==> '(?,)'
The (?,) means that y is a vector of unknown length. Why is it unknown? tf.unique(x) returns the unique values from x, and the values of x are unknown because it is a tf.placeholder(), so it doesn’t have a value until you feed it. Let’s see what happens if you feed two different values:
sess = tf.Session()
print sess.run(y, feed_dict={x: [0, 1, 2, 3]}).shape
# ==> '(4,)'
print sess.run(y, feed_dict={x: [0, 0, 0, 0]}).shape
# ==> '(1,)'
Hopefully this makes it clear that a tensor can have a different static and dynamic shape. The dynamic shape is always fully defined—it has no ? dimensions—but the static shape can be less specific. This is what allows TensorFlow to support operations like tf.unique() and tf.dynamic_partition(), which can have variable-sized outputs, and are used in advanced applications.
Finally, the tf.shape() op can be used to get the dynamic shape of a tensor and use it in a TensorFlow computation:
z = tf.shape(y)
print sess.run(z, feed_dict={x: [0, 1, 2, 3]})
# ==> [4]
print sess.run(z, feed_dict={x: [0, 0, 0, 0]})
# ==> [1]
Here’s a schematic image showing both:
