/TensorFlow 2.4


Normalizes along dimension axis using an L2 norm.

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.

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).
A Tensor with the same shape as x.

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Code samples licensed under the Apache 2.0 License.