Take a look at the following code sample:
def myFun(my_tensor):
#The following line works
my_tensor= tf.tensor_scatter_update(my_tensor, tf.constant([[0]]), tf.constant([1]))
#The following line leads to error
p = tf.cond(tf.math.equal(0, 0), lambda: 1, lambda: 1)
my_tensor= tf.tensor_scatter_update(my_tensor, tf.constant([[p]]), tf.constant([1]))
I have taken a simple case to describe the issue I am facing This function (myFun) is called as the body of a tf.while_loop (in case that is relevant) Definiton of my_tensor
my_tensor = tf.zeros(5, tf.int32)
How do I define the indices parameter of the tf.tensor_scatter_update? I am using tensorflow1.15
CodePudding user response:
You cannot use the tensor p
as an argument for tf.constant
. Maybe try something like this:
%tensorflow_version 1.x
import tensorflow as tf
def myFun(my_tensor):
my_tensor= tf.tensor_scatter_update(my_tensor, tf.constant([[0]]), tf.constant([1]))
p = tf.cond(tf.math.equal(0, 0), lambda: 1, lambda: 1)
new_tensor= tf.tensor_scatter_update(my_tensor, [[p]], tf.constant([1]))
with tf.Session() as sess:
p_value = p.eval()
tensor_values = my_tensor.eval()
new_tensor_values = new_tensor.eval()
print('p -->', p_value)
print('my_tensor -->', tensor_values)
print('new_tensor -->', new_tensor_values)
my_tensor = tf.zeros(5, tf.int32)
myFun(my_tensor)
p --> 1
my_tensor --> [1 0 0 0 0]
new_tensor --> [1 1 0 0 0]
You can also wrap p
around a tf.Variable
:
def myFun(my_tensor):
my_tensor= tf.tensor_scatter_update(my_tensor, tf.constant([[0]]), tf.constant([1]))
p = tf.cond(tf.math.equal(0, 0), lambda: 1, lambda: 1)
indices = tf.Variable([[p]])
new_tensor= tf.tensor_scatter_update(my_tensor, indices, tf.constant([1]))
with tf.Session() as sess:
sess.run(indices.initializer)
p_value = p.eval()
tensor_values = my_tensor.eval()
new_tensor_values = new_tensor.eval()