Tensorflow:从张量堆叠所有行对

Question

鉴于张量t=[[1,2], [3,4]],我需要生产ts=[[1,2,1,2], [1,2,3,4], [3,4,1,2], [3,4,3,4]].也就是说,我需要将所有行对堆叠在一起.重要:张量具有维度[无,2],即.第一个维度是可变的.

我试过了:

• 使用tf.while_loop生成的索引列表idx=[[0, 0], [0, 1], [1, 0], [1, 1]],然后tf.gather(ts, idx).这有效,但很乱,我不知道如何处理渐变.
• 2 for循环遍历tf.unstack(t),将堆叠的行添加到缓冲区,然后tf.stack(buffer).如果第一个维度是可变的,则这不起作用.
• 寻找广播的灵感.例如,给定的x=t.expand_dims(t, 0), y=t.expand_dims(t, 1), s=tf.reshape(tf.add(x, y), [-1, 2]) s是[[2,4],[4,6],[4,6],[6,8]],即.每行组合的总和.但是我怎样才能进行堆叠而不是求和？我已经失败了2天:)

Answer 1

解决方案tf.meshgrid()和一些重塑：

import tensorflow as tf
import numpy as np

t = tf.placeholder(tf.int32, [None, 2])
num_rows, size_row = tf.shape(t)[0], tf.shape(t)[1] # actual dynamic dimensions

# Getting pair indices using tf.meshgrid:
idx_range = tf.range(num_rows)
pair_indices = tf.stack(tf.meshgrid(*[idx_range, idx_range]))
pair_indices = tf.transpose(pair_indices, perm=[1, 2, 0])

# Finally gathering the rows accordingly:
res = tf.reshape(tf.gather(t, pair_indices), (-1, size_row * 2))

with tf.Session() as sess:
    print(sess.run(res, feed_dict={t: np.array([[1,2], [3,4], [5,6]])}))
    # [[1 2 1 2]
    #  [3 4 1 2]
    #  [5 6 1 2]
    #  [1 2 3 4]
    #  [3 4 3 4]
    #  [5 6 3 4]
    #  [1 2 5 6]
    #  [3 4 5 6]
    #  [5 6 5 6]]

使用笛卡尔积的解决方案：

import tensorflow as tf
import numpy as np

t = tf.placeholder(tf.int32, [None, 2])
num_rows, size_row = tf.shape(t)[0], tf.shape(t)[1] # actual dynamic dimensions

# Getting pair indices by computing the indices cartesian product:
row_idx = tf.range(num_rows)
row_idx_a = tf.expand_dims(tf.tile(tf.expand_dims(row_idx, 1), [1, num_rows]), 2)
row_idx_b = tf.expand_dims(tf.tile(tf.expand_dims(row_idx, 0), [num_rows, 1]), 2)
pair_indices = tf.concat([row_idx_a, row_idx_b], axis=2)

# Finally gathering the rows accordingly:
res = tf.reshape(tf.gather(t, pair_indices), (-1, size_row * 2))

with tf.Session() as sess:
    print(sess.run(res, feed_dict={t: np.array([[1,2], [3,4], [5,6]])}))
    # [[1 2 1 2]
    #  [1 2 3 4]
    #  [1 2 5 6]
    #  [3 4 1 2]
    #  [3 4 3 4]
    #  [3 4 5 6]
    #  [5 6 1 2]
    #  [5 6 3 4]
    #  [5 6 5 6]]