tf.math.l2_normalize

Normalizes along dimension axis using an L2 norm. (deprecated arguments)

View aliases

Main aliases

tf.linalg.l2_normalize, tf.nn.l2_normalize

Compat aliases for migration

See Migration guide for more details.

tf.compat.v1.linalg.l2_normalize, tf.compat.v1.math.l2_normalize, tf.compat.v1.nn.l2_normalize

tf.math.l2_normalize(
    x, axis=None, epsilon=1e-12, name=None, dim=None
)

Deprecated: SOME ARGUMENTS ARE DEPRECATED: (dim). They will be removed in a future version. Instructions for updating: dim is deprecated, use axis instead

For a 1-D tensor with axis = 0, computes

output = x / sqrt(max(sum(x**2), epsilon))

For x with more dimensions, independently normalizes each 1-D slice along dimension axis.

1-D tensor example:

>>> x = tf.constant([3.0, 4.0])
>>> tf.math.l2_normalize(x).numpy()
array([0.6, 0.8], dtype=float32)

2-D tensor example:

>>> x = tf.constant([[3.0], [4.0]])
>>> tf.math.l2_normalize(x, 0).numpy()
array([[0.6],
     [0.8]], dtype=float32)

x = tf.constant([[3.0], [4.0]])
tf.math.l2_normalize(x, 1).numpy()
array([[1.],
     [1.]], dtype=float32)

Args
x	A Tensor.
axis	Dimension along which to normalize. A scalar or a vector of integers.
epsilon	A lower bound value for the norm. Will use sqrt(epsilon) as the divisor if norm < sqrt(epsilon).
name	A name for this operation (optional).
dim	Deprecated, do not use.

Returns
A Tensor with the same shape as x.