tf.shape  |  TensorFlow v2.16.1 (original) (raw)

tf.shape

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Returns a tensor containing the shape of the input tensor.

tf.shape(
    input, out_type=None, name=None
)

Used in the notebooks

See also tf.size, tf.rank.

tf.shape returns a 1-D integer tensor representing the shape of input. For a scalar input, the tensor returned has a shape of (0,) and its value is the empty vector (i.e. []).

For example:

tf.shape(1.) <tf.Tensor: shape=(0,), dtype=int32, numpy=array([], dtype=int32)>

t = tf.constant([[[1, 1, 1], [2, 2, 2]], [[3, 3, 3], [4, 4, 4]]]) tf.shape(t) <tf.Tensor: shape=(3,), dtype=int32, numpy=array([2, 2, 3], dtype=int32)>

a = tf.keras.layers.Input((None, 10)) tf.shape(a) <... shape=(3,) dtype=int32...>

In these cases, using tf.Tensor.shape will return more informative results.

a.shape TensorShape([None, None, 10])

(The first None represents the as yet unknown batch size.)

tf.shape and Tensor.shape should be identical in eager mode. Withintf.function or within a compat.v1 context, not all dimensions may be known until execution time. Hence, when defining custom layers and models for graph mode, prefer the dynamic tf.shape(x) over the static x.shape.