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 sha

Answer 1

import tensorflow as tf

def broadcast(tensor, shape):
     """Broadcasts ``x`` to have shape ``shape``.
                                                                   |
     Uses ``tf.Assert`` statements to ensure that the broadcast is
     valid.

     First calculates the number of missing dimensions in 
     ``tf.shape(x)`` and left-pads the shape of ``x`` with that many 
     ones. Then identifies the dimensions of ``x`` that require
     tiling and tiles those dimensions appropriately.

     Args:
         x (tf.Tensor): The tensor to broadcast.
         shape (Union[tf.TensorShape, tf.Tensor, Sequence[int]]): 
             The shape to broadcast to.

     Returns:
         tf.Tensor: ``x``, reshaped and tiled to have shape ``shape``.

     """
     with tf.name_scope('broadcast') as scope:
         shape_x = tf.shape(x)
         rank_x = tf.shape(shape0)[0]
         shape_t = tf.convert_to_tensor(shape, preferred_dtype=tf.int32)
         rank_t = tf.shape(shape1)[0]

         with tf.control_dependencies([tf.Assert(
             rank_t >= rank_x,
             ['len(shape) must be >= tf.rank(x)', shape_x, shape_t],
             summarize=255
         )]):
             missing_dims = tf.ones(tf.stack([rank_t - rank_x], 0), tf.int32)

         shape_x_ = tf.concat([missing_dims, shape_x], 0)
         should_tile = tf.equal(shape_x_, 1)

         with tf.control_dependencies([tf.Assert(
             tf.reduce_all(tf.logical_or(tf.equal(shape_x_, shape_t), should_tile),
             ['cannot broadcast shapes', shape_x, shape_t],
             summarize=255
         )]):
             multiples = tf.where(should_tile, shape_t, tf.ones_like(shape_t))
             out = tf.tile(tf.reshape(x, shape_x_), multiples, name=scope)

         try:
             out.set_shape(shape)
         except:
             pass

         return out

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)

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)

Answer 3

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.

---

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 4

In the newest tensorflow version(2.0), you can use tf.broadcast_to as below:

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.broadcast_to(C, A.shape)

D = tf.where(C,A,B)