TensorFlow concat a variable-sized placeholder with a vector

Question

Let's say I have a placeholder

ph_input = tf.placeholder(dtype=tf.int32, [None, 1])

and a vector

h = tf.zeros([1,2], dtype=tf.int32)

In this example h is filled with zeros for simplicity but in the actual case it will be changed by other variables and will have different values.

I want to efficiently do a concat on ph_input and h on dimension 1 and get a new tensor with shape [None, 1+2]. Unfortunately concat needs all input tensors to have the same shape except the concat_dim, which my example does not meet.

I was considering expand h to the same shape as the data that feed to ph_input but am not exactly sure how to do that with the placeholder itself. If I get the shape from the input data directly then I guess it is not necessary to use the placeholder.

Answer 1

The most general solution is to use the tf.shape() op to get the run-time size of the placeholder, and the tf.tile() op to expand h to the appropriate size:

ph_input = tf.placeholder(dtype=tf.int32, shape=[None, 1])
h = tf.zeros([1, 2], dtype=tf.int32)  # ...or some other tensor of shape [1, 2]

# Get the number of rows in the fed value at run-time.
ph_num_rows = tf.shape(ph_input)[0]

# Makes a `ph_num_rows x 2` matrix, by tiling `h` along the row dimension.
h_tiled = tf.tile(h, tf.pack([ph_num_rows, 1]))

result = tf.concat(1, [ph_input, h_tiled])