评估tensorflow中两个张量的行的所有对组合

Question

我试图在tensorflow中定义一个自定义op,其中我需要构造一个矩阵(z),它将包含两个矩阵(x和y)的行对的所有组合的总和.一般情况下,行数x和y有动力.

在numpy中它很简单:

import numpy as np
from itertools import product

rows_x = 4
rows_y = 2
dim = 2

x = np.arange(dim*rows_x).reshape(rows_x, dim)
y = np.arange(dim*rows_y).reshape(rows_y, dim)

print('x:\n{},\ny:\n{}\n'.format(x, y))

z = np.zeros((rows_x*rows_y, dim))
print('for loop:')
for i, (x_id, y_id) in enumerate(product(range(rows_x), range(rows_y))):
    print('row {}: {} + {}'.format(i, x[x_id, ], y[y_id, ]))
    z[i, ] = x[x_id, ] + y[y_id, ]

print('\nz:\n{}'.format(z))

收益:

x:
[[0 1]
 [2 3]
 [4 5]
 [6 7]],
y:
[[0 1]
 [2 3]]

for loop:
row 0: [0 1] + [0 1]
row 1: [0 1] + [2 3]
row 2: [2 3] + [0 1]
row 3: [2 3] + [2 3]
row 4: [4 5] + [0 1]
row 5: [4 5] + [2 3]
row 6: [6 7] + [0 1]
row 7: [6 7] + [2 3]

z:
[[  0.   2.]
 [  2.   4.]
 [  2.   4.]
 [  4.   6.]
 [  4.   6.]
 [  6.   8.]
 [  6.   8.]
 [  8.  10.]]

但是,我还没有弄清楚如何在tensorflow中实现类似的任何东西.

我主要是通过SO和tensorflow API来寻找能够产生两个张量元素组合的函数,或者是一个能够给出张量元素排列的函数,但无济于事.

任何建议都是最受欢迎的.

Answer 1

你可以简单地使用张量流的广播能力.

import tensorflow as tf

x = tf.constant([[0, 1],[2, 3],[4, 5],[6, 7]], dtype=tf.float32)
y = tf.constant([[0, 1],[2, 3]], dtype=tf.float32)

x_ = tf.expand_dims(x, 0)
y_ = tf.expand_dims(y, 1)
z = tf.reshape(tf.add(x_, y_), [-1, 2])
# or more succinctly 
z = tf.reshape(x[None] + y[:, None], [-1, 2])

sess = tf.Session()
sess.run(z)