构造tf.variable对象的张量

Question

My cost function involves the matrix

T=[[1.0-a,b],[a,1.0-b]] 

I can define

import numpy as np
import tensorflow as tf
a=0.3
b=0.4
T = tf.Variable([[1.0-a,b],[a,1.0-b]]

and this works well in the optimization, but then I am saying that I have four variables: 1-a,b,a,1-b (the gradient has four elements). On the other hand, I would like my variables to be two: a and b (the gradient has two elements).

I thought of doing something like

var = tf.Variable([a,b])
T = tf.constant([[1.0-var[0],var[1]],[var[0],1.0-var[1]]])

but this does not work, outputing the following:

TypeError: List of Tensors when single Tensor expected

So how can I construct a tensor made of tf.Variable objects?

Thank you.

Answer 1

I think what you need is:

import tensorflow as tf

a = tf.Variable(0.3)
b = tf.Variable(0.4)
T = tf.convert_to_tensor([[1.0 - a, b], [a, 1.0 - b]]

Answer 2

Initialise T as a placeholder.

T = tf.placeholder(tf.float32, [2, 2])

When starting your session, compute T and pass it in through a feed_dict :

with tf.Session() as sess:
    a, b = .3, .4
    inp = np.array([[1 - a, b], [a, 1 - b]])
    sess.run(optimizer, feed_dict={T : inp})

Where optimizer is the node that minimizes your cost function.