|  TensorFlow 1 version |  View source on GitHub | 
Normalizes along dimension axis using an L2 norm.
tf.math.l2_normalize(
    x, axis=None, epsilon=1e-12, name=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.
| 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 ifnorm < sqrt(epsilon). | 
| name | A name for this operation (optional). | 
| Returns | |
|---|---|
| A Tensorwith the same shape asx. |