reduce_sum by certain dimension

Question

I have two embeddings tensor A and B, which looks like

[
  [1,1,1],
  [1,1,1]
]

and

[
  [0,0,0],
  [1,1,1]
]

what I want to do is calculate the L2 distance d(A,B) element-wise.

First I did a tf.square(tf.sub(lhs, rhs)) to get

[
  [1,1,1],
  [0,0,0]
]

and then I want to do an element-wise reduce which returns

[
  3,
  0
]

but tf.reduce_sum does not allow my to reduce by row. Any inputs would be appreciated. Thanks.

Answer 1

Add the reduction_indices argument with a value of 1, eg.:

tf.reduce_sum( tf.square( tf.sub( lhs, rhs) ), 1 )

That should produce the result you're looking for. Here is the documentation on reduce_sum().

Answer 2

According to TensorFlow documentation, reduce_sum function which takes four arguments.

tf.reduce_sum(input_tensor, axis=None, keep_dims=False, name=None, reduction_indices=None).

But reduction_indices has been deprecated. Better to use axis instead of. If the axis is not set, reduces all its dimensions.

As an example,this is taken from the documentation,

# 'x' is [[1, 1, 1]
#         [1, 1, 1]]
tf.reduce_sum(x) ==> 6
tf.reduce_sum(x, 0) ==> [2, 2, 2]
tf.reduce_sum(x, 1) ==> [3, 3]
tf.reduce_sum(x, 1, keep_dims=True) ==> [[3], [3]]
tf.reduce_sum(x, [0, 1]) ==> 6

Above requirement can be written in this manner,

import numpy as np
import tensorflow as tf

a = np.array([[1,7,1],[1,1,1]])
b = np.array([[0,0,0],[1,1,1]])

xtr = tf.placeholder("float", [None, 3])
xte = tf.placeholder("float", [None, 3])

pred = tf.reduce_sum(tf.square(tf.subtract(xtr, xte)),1)

# Initializing the variables
init = tf.global_variables_initializer()

# Launch the graph
with tf.Session() as sess:
    sess.run(init)
    nn_index = sess.run(pred, feed_dict={xtr: a, xte: b})
    print nn_index