Cartesian Product in Tensorflow

I'm going to assume here that both a and b are 1-D tensors.

To get the cartesian product of the two, I would use a combination of tf.expand_dims and tf.tile:

a = tf.constant([1,2,3]) 
b = tf.constant([4,5,6,7]) 

tile_a = tf.tile(tf.expand_dims(a, 1), [1, tf.shape(b)[0]])  
tile_a = tf.expand_dims(tile_a, 2) 
tile_b = tf.tile(tf.expand_dims(b, 0), [tf.shape(a)[0], 1]) 
tile_b = tf.expand_dims(tile_b, 2) 

cartesian_product = tf.concat([tile_a, tile_b], axis=2) 

cart = tf.Session().run(cartesian_product) 

print(cart.shape) 
print(cart) 

You end up with a len(a) * len(b) * 2 tensor where each combination of the elements of a and b is represented in the last dimension.

A shorter solution to the same, using tf.add() for broadcasting (tested):

import tensorflow as tf

a = tf.constant([1,2,3]) 
b = tf.constant([4,5,6,7]) 

a, b = a[ None, :, None ], b[ :, None, None ]
cartesian_product = tf.concat( [ a + tf.zeros_like( b ),
                                 tf.zeros_like( a ) + b ], axis = 2 )

with tf.Session() as sess:
    print( sess.run( cartesian_product ) )

will output:

[[[1 4]
[2 4]
[3 4]]

[[1 5]
[2 5]
[3 5]]

[[1 6]
[2 6]
[3 6]]

[[1 7]
[2 7]
[3 7]]]

import tensorflow as tf

a = tf.constant([0, 1, 2])
b = tf.constant([2, 3])
c = tf.stack(tf.meshgrid(a, b, indexing='ij'), axis=-1)
c = tf.reshape(c, (-1, 2))
with tf.Session() as sess:
    print(sess.run(c))

Output:

[[0 2]
 [0 3]
 [1 2]
 [1 3]
 [2 2]
 [2 3]]

credit to jdehesa: link

A more succinct version of Sunreef's answer uses tf.stack instead of tf.concat

a = tf.constant([1,2,3]) 
b = tf.constant([4,5,6,7]) 

tile_a = tf.tile(tf.expand_dims(a, 1), [1, tf.shape(b)[0]]) 
tile_b = tf.tile(tf.expand_dims(b, 0), [tf.shape(a)[0], 1])  
ans = tf.stack([tile_a, tile_b], -1)