CODEWITHSUNDEEP
tensorflow
mund_ak
mund_ak

Reputation: 133

Evaluate a condition on each element of a vector y in tensorflow

I am trying to evaluate a condition on each element of a vector y so that I get a vector whose i’th element tells me whether y[i]satisfies the condition. Is there any way to do this without using loops? So far, I have tried the following:

dim = 3

x = tf.placeholder(tf.float32, shape = [dim])
y = tf.log(x)

tf1 = tf.constant(1)
tf0 = tf.constant(0)


x_0 = tf.tile([x[0]], [dim])

delta = tf.cond(tf.equal(y,x_0),  tf1, tf0))
sess = tf.Session()
a = np.ones((1,3))

print(sess.run(delta, feed_dict={x:a}))

For a given input x, I want delta[i] to be 1 if y[i] = x[0] and 0 otherwise.

I get error

shape must be of equal rank but are 0 and 1 for 'Select_2' (op: 'select') with input shapes [3], [],[]

I am new to TensorFlow, any help would be appreciated!

Upvotes: 3

Views: 615

Answers (2)

V. Smirnov
V. Smirnov

Reputation: 65

Seems like that you have error because you are trying to compare tensors with different shape.

That's working code:

import tensorflow as tf
import numpy as np

dim = 3

x = tf.placeholder(tf.float32, shape=(1, dim), name='ktf')
y = tf.log(x)

delta = tf.cast(tf.equal(y, x[0]), dtype=tf.int32)

sess = tf.Session()
a = np.ones((1, 3))

print(sess.run(delta, feed_dict={x: a}))

Upvotes: 1

Cindy Almighty
Cindy Almighty

Reputation: 933

For you case, there is no need to use tf.cond, you can use tf.equal that does this without the loops, and because of the broadcasting there is no need to tile it. Just use:

dim = 3

x = tf.placeholder(tf.float32, shape = [dim])
y = tf.log(x)

delta = tf.cast(tf.equal(y,x[0]),tf.float32) # or integer type

sess = tf.Session()
a = np.ones((1,3))

print(sess.run(delta, feed_dict={x:a}))

Upvotes: 0

Related Questions