Python – tensorflow.cond()

TensorFlow is open-source Python library designed by Google to develop Machine Learning models and deep learning  neural networks.

cond() return true_fn() if the predicate pred is true otherwise it returns false_fn().

Syntax: tensorflow.cond( pred, true_fn, false_fn, name )

Parameters:

  • pred: It is a scalar which determines the callable to return
  • true_fn(optional): It is returned when pred is true.
  • false_fn(optional): It is returned when pred is false.
  • name(optional): It defines the name for the operation.

Return: It returns the result evaluated by callable.



Example 1:

Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Importing the library
import tensorflow as tf
  
# Initializing the input
x = 5
y = 10
  
  
# Printing the input
print('x: ', x)
print('y: ', y)
  
# Calculating result
res = tf.cond(x < y, lambda: tf.add(x, y), lambda: tf.square(y))
  
# Printing the result
print('Result: ', res)

chevron_right


Output:

x:  5
y:  10
Result:  tf.Tensor(15, shape=(), dtype=int32)

Example 2:

Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Importing the library
import tensorflow as tf
  
# Initializing the input
x = 5
y = 10
  
  
# Printing the input
print('x: ', x)
print('y: ', y)
  
# Calculating result
res = tf.cond(x > y, lambda: tf.add(x, y), lambda: tf.square(y))
  
# Printing the result
print('Result: ', res)

chevron_right


Output:

x:  5
y:  10
Result:  tf.Tensor(100, shape=(), dtype=int32)



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.


Article Tags :

Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.