Question

我有一个嵌入矩阵，并且有一个3D稀疏张量用于获取嵌入输出，在阅读tf.nn.embedding_lookup_sparse的文档后，我发现它仅支持2D稀疏张量，

sp_ids：N x M个int64 id的SparseTensor，其中N通常是批处理大小，M是任意的。

我的示例代码在这里

import numpy as np
import tensorflow as tf
tf.enable_eager_execution()

# [feature number, embedding dim] 
w = tf.get_variable("w", [4, 4], initializer=tf.random_normal_initializer())

z = np.array(
     [
      [
        [0, 1, 2, 3],   # get the vector of row 0, 1, 2, 3 of the embedding matrix w and get the sum
        [2, 3]
      ],

      [
        [1, 3],
        [2]
      ],

      [
        [0, 1, 3],
        [1, 2]
      ]
     ])

sp = tf.SparseTensor(values=[0, 1, 2, 3, 2, 3, 1, 3, 2, 0, 1, 3, 1, 2],
                     indices=[[0,0,0],[0,0,1],[0,0,2],[0,0,3],[0,1,2],
                              [0,1,3],[1,0,1],[1,0,3],[1,1,2],[2,0,0],
                              [2,0,1],[2,0,3],[2,1,1],[2,1,2]],
                     dense_shape=[3, 2, 4])

tf.nn.embedding_lookup_sparse(w, sp, None, combiner='sum')
# the outputs
<tf.Tensor: id=970, shape=(3, 4), dtype=float32, numpy=
array([[-5.8729677 , -1.3900641 ,  0.8126096 , -3.1223912 ],
       [-1.0788026 , -1.1324122 ,  0.34160078,  0.23714277],
       [-2.497394  , -2.7855003 ,  3.0201516 , -1.8009453 ]],
      dtype=float32)>

print(w)
<tf.Variable 'w:0' shape=(4, 4) dtype=float32, numpy=
array([[-2.5669768 , -0.38916406,  1.4039794 , -2.8173826 ],
       [ 1.1483854 , -1.2639242 ,  1.2745714 ,  0.7792944 ],
       [-1.3400027 , -0.46362385, -1.3652185 ,  0.27220532],
       [-0.8871854 ,  0.5951359 ,  0.43224794, -0.8143569 ]],
      dtype=float32)>

但是预期输出是尺寸为3x2x4而不是3x4的矩阵。 tf.nn.embedding_lookup_sparse是否支持此操作？

Answer 1

最简单的方法是将稀疏张量设为2D张量，并获得嵌入矩阵的权重，然后重塑。

# First make the z as a 2D arr and create a sparse tensor 
z = np.array([
        [0, 1, 2, 3],  # get the row 0,1,2,3 of the embedding matrix w and get the sum
        [2, 3],
        [1, 3],
        [2],
        [0, 1, 3],
        [1, 2]
      ])

sp = tf.SparseTensor(values=[0, 1, 2, 3, 2, 3, 1, 3, 2, 0, 1, 3, 1, 2],
                     indices=[[0,0],[0,1],[0,2],[0,3],[1,2],[1,3],[2,1],
                              [2,3],[3,2],[4,0],[4,1],[4,3],[5,1],[5,2]],
                     dense_shape=[6, 4])

res = tf.nn.embedding_lookup_sparse(w, sp, None, combiner='sum')

res.numpy()
# the output
array([[-3.6457794 , -1.5215762 ,  1.7455802 , -2.5802398 ],
       [-2.227188  ,  0.13151208, -0.9329706 , -0.5421516 ],
       [ 0.2612    , -0.6687883 ,  1.7068193 , -0.03506255],
       [-1.3400027 , -0.46362385, -1.3652185 ,  0.27220532],
       [-2.3057768 , -1.0579524 ,  3.1107986 , -2.8524451 ],
       [-0.19161725, -1.7275481 , -0.0906471 ,  1.0514997 ]],

# reshape
tf.reshape(res, [-1, 2, 4])
# that is exacly what I want.
array([[[-3.6457794 , -1.5215762 ,  1.7455802 , -2.5802398 ],
        [-2.227188  ,  0.13151208, -0.9329706 , -0.5421516 ]],

       [[ 0.2612    , -0.6687883 ,  1.7068193 , -0.03506255],
        [-1.3400027 , -0.46362385, -1.3652185 ,  0.27220532]],

       [[-2.3057768 , -1.0579524 ,  3.1107986 , -2.8524451 ],
        [-0.19161725, -1.7275481 , -0.0906471 ,  1.0514997 ]]])

# print w, and the above result is right
w.numpy()

array([[-2.5669768 , -0.38916406,  1.4039794 , -2.8173826 ],
       [ 1.1483854 , -1.2639242 ,  1.2745714 ,  0.7792944 ],
       [-1.3400027 , -0.46362385, -1.3652185 ,  0.27220532],
       [-0.8871854 ,  0.5951359 ,  0.43224794, -0.8143569 ]],
      dtype=float32)

因此，忘记3D稀疏张量，只需将其转换为2D张量即可。因为您只关心稀疏张量中的值（行索引，用于获取嵌入矩阵的相应行）。

tf.nn.embedding_lookup_sparse 3D稀疏张量输入

1 个答案: