tf.math.l2_normalize
    
    
       
    
    
      
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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 
) 
 
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.
 
 
  
  
 
  
    
    
      
    
     
  
       
    
    
  
  
 
  Except as otherwise noted, the content of this page is licensed under the Creative Commons Attribution 4.0 License , and code samples are licensed under the Apache 2.0 License . For details, see the Google Developers Site Policies . Java is a registered trademark of Oracle and/or its affiliates. Some content is licensed under the numpy license .
  Last updated 2023-10-06 UTC.
 
 
  
  
  
    
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