Use tf.scatter_update in a two dimensional tf.Variable

Question

I' m following this Manipulating matrix elements in tensorflow. using tf.scatter_update. But my problem is: What happens if my tf.Variable is 2D? Let's say:

a = tf.Variable(initial_value=[[0, 0, 0, 0],[0, 0, 0, 0]])

How can i update for example the first element of every row and assign to that the value 1?

I tried something like

for line in range(2):
    sess.run(tf.scatter_update(a[line],[0],[1]))

but it fails (i was expecting that) and gives me the error:

TypeError: Input 'ref' of 'ScatterUpdate' Op requires l-value input

How can i fix that kind of problems?

`

score 14 · Accepted Answer · answered Oct 29 '16 at 17:46

In tensorflow you cannot update a Tensor but you can update a Variable.

The scatter_update operator can update only the first dimension of the variable. You have to pass always a reference tensor to the scatter update (a instead of a[line]).

This is how you can update the first element of the variable:

import tensorflow as tf

g = tf.Graph()
with g.as_default():
    a = tf.Variable(initial_value=[[0, 0, 0, 0],[0, 0, 0, 0]])
    b = tf.scatter_update(a, [0, 1], [[1, 0, 0, 0], [1, 0, 0, 0]])

with tf.Session(graph=g) as sess:
   sess.run(tf.initialize_all_variables())
   print sess.run(a)
   print sess.run(b)

Output:

[[0 0 0 0]
 [0 0 0 0]]
[[1 0 0 0]
 [1 0 0 0]]

But having to change again the whole tensor it might be faster to just assign a completely new one.

Note that, after the `tf.scatter_update` operation, the value of `a` will also change and will be equal to `b`. — Ataxias, Mar 20 '19 at 03:01
Also, both `a` and `b` will become Tensors and will no longer be of type `Variable`. — Ataxias, Mar 22 '19 at 00:06

score 1 · Answer 2 · answered May 07 '18 at 05:35

I found something here I made a variable name U = [[1, 2, 3], [4, 5, 6]] and wanted to update it like as U[:,1] = [2, 3] so I did U[:,1].assign(cast_into_tensor[2,3])

here a simple code

x = tf.Variable([[1,2,3],[4,5,6]])
print K.eval(x)
y = [0, 0]
with tf.control_dependencies([x[:,1].assign(y)]):
    x = tf.identity(x)
print K.eval(x)

Use tf.scatter_update in a two dimensional tf.Variable

2 Answers2