How to explicitly broadcast a tensor to match another's shape in tensorflow?

Question

I have three tensors, A, B and C in tensorflow, A and B are both of shape (m, n, r), C is a binary tensor of shape (m, n, 1).

I want to select elements from either A or B based on the value of C. The obvious tool is tf.select, however that does not have broadcasting semantics, so I need to first explicitly broadcast C to the same shape as A and B.

This would be my first attempt at how to do this, but it doesn't like me mixing a tensor (tf.shape(A)[2]) into the shape list.

import tensorflow as tf
A = tf.random_normal([20, 100, 10])
B = tf.random_normal([20, 100, 10])
C = tf.random_normal([20, 100, 1])
C = tf.greater_equal(C, tf.zeros_like(C))

C = tf.tile(C, [1,1,tf.shape(A)[2]])
D = tf.select(C, A, B)

What's the correct approach here?

Answer 1

EDIT: In all versions of TensorFlow since 0.12rc0, the code in the question works directly. TensorFlow will automatically stack tensors and Python numbers into a tensor argument. The solution below using tf.pack() is only needed in versions prior to 0.12rc0. Note that tf.pack() was renamed to tf.stack() in TensorFlow 1.0.

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Your solution is very close to working. You should replace the line:

C = tf.tile(C, [1,1,tf.shape(C)[2]])

...with the following:

C = tf.tile(C, tf.pack([1, 1, tf.shape(A)[2]]))

(The reason for the issue is that TensorFlow won't implicitly convert a list of tensors and Python literals into a tensor. tf.pack() takes a list of tensors, so it will convert each of the elements in its input (1, 1, and tf.shape(C)[2]) to a tensor. Since each element is a scalar, the result will be a vector.)

Answer 2

Here's a dirty hack:

import tensorflow as tf

def broadcast(tensor, shape):
    return tensor + tf.zeros(shape, dtype=tensor.dtype)

A = tf.random_normal([20, 100, 10])
B = tf.random_normal([20, 100, 10])
C = tf.random_normal([20, 100, 1])

C = broadcast(C, A.shape)
D = tf.select(C, A, B)