How does reduce_sum() work in tensorflow?

Question

I am learning tensorflow, I picked up the following code from the tensorflow website. According to my understanding, axis=0 is for rows and axis=1 is for columns.

How are they getting output mentioned in comments? I have mentioned output according to my thinking against ##.

import tensorflow as tf  x = tf.constant([[1, 1, 1], [1, 1, 1]]) tf.reduce_sum(x, 0)  # [2, 2, 2] ## [3, 3] tf.reduce_sum(x, 1)  # [3, 3] ##[2, 2, 2] tf.reduce_sum(x, [0, 1])  # 6 ## Didn't understand at all. 

Answer 1

x has a shape of (2, 3) (two rows and three columns):

1 1 1 1 1 1 

By doing tf.reduce_sum(x, 0) the tensor is reduced along the first dimension (rows), so the result is [1, 1, 1] + [1, 1, 1] = [2, 2, 2].

By doing tf.reduce_sum(x, 1) the tensor is reduced along the second dimension (columns), so the result is [1, 1] + [1, 1] + [1, 1] = [3, 3].

By doing tf.reduce_sum(x, [0, 1]) the tensor is reduced along BOTH dimensions (rows and columns), so the result is 1 + 1 + 1 + 1 + 1 + 1 = 6 or, equivalently, [1, 1, 1] + [1, 1, 1] = [2, 2, 2], and then 2 + 2 + 2 = 6 (reduce along rows, then reduce the resulted array).

Answer 2

In order to understand better what is going on I will change the values, and the results are self explanatory

import tensorflow as tf  x = tf.constant([[1, 2, 4], [8, 16, 32]]) a = tf.reduce_sum(x, 0)  # [ 9 18 36] b = tf.reduce_sum(x, 1)  # [ 7 56] c = tf.reduce_sum(x, [0, 1])  # 63  with tf.Session() as sess:   output_a = sess.run(a)   print(output_a)   output_b = sess.run(b)   print(output_b)   output_c = sess.run(c)   print(output_c)