tf.math.l2_normalize Stay organized with collections Save and categorize content based on your preferences.
Normalizes along dimension axis using an L2 norm. (deprecated arguments)
tf . math . l2_normalize ( x , axis = None , epsilon = 1e-12 , name = None , dim = None ) Used in the notebooks 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.
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Last updated 2024-04-26 UTC.
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