TensorFlow 1 version
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    View source on GitHub
  
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Normalizes along dimension axis using an L2 norm. (deprecated arguments)
tf.math.l2_normalize(
    x, axis=None, epsilon=1e-12, name=None, dim=None
)
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.
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  TensorFlow 1 version
    View source on GitHub