python - 在 tensorflow 中广播 SparseTensor

Question

我想将具有 shape 的密集张量与具有 shape[n, n, k]的稀疏张量相乘[n, n, 1]。我希望稀疏张量中的值以 size 沿轴重复s，就像我使用密集张量并依赖隐式广播一样。

但是该SparseTensor.__mul__操作不支持广播稀疏操作数。我没有找到明确广播稀疏张量的运算符。我怎么能做到这一点？

标签： pythontensorflow

Answer 1

如果您不想只是将稀疏张量转换为密集，您可以从密集张量中提取选择正确的值以直接构建稀疏结果，如下所示：

import tensorflow as tf
import numpy as np

with tf.Graph().as_default(), tf.Session() as sess:
    # Input data
    x = tf.placeholder(tf.float32, shape=[None, None, None])
    y = tf.sparse.placeholder(tf.float32, shape=[None, None, 1])
    # Indices of sparse tensor without third index coordinate
    indices2 = y.indices[:, :-1]
    # Values of dense tensor corresponding to sparse tensor values
    x_sp = tf.gather_nd(x, indices2)
    # Values of the resulting sparse tensor
    res_vals = tf.reshape(x_sp * tf.expand_dims(y.values, 1), [-1])
    # Shape of the resulting sparse tensor
    res_shape = tf.shape(x, out_type=tf.int64)
    # Make sparse tensor indices
    k = res_shape[2]
    v = tf.size(y.values)
    # Add third coordinate to existing sparse tensor coordinates
    idx1 = tf.tile(tf.expand_dims(indices2, 1), [1, k, 1])
    idx2 = tf.tile(tf.range(k), [v])
    res_idx = tf.concat([tf.reshape(idx1, [-1, 2]), tf.expand_dims(idx2, 1)], axis=1)
    # Make sparse result
    res = tf.SparseTensor(res_idx, res_vals, res_shape)
    # Dense value for testing
    res_dense = tf.sparse.to_dense(res)
    # Dense operation for testing
    res_dense2 = x * tf.sparse.to_dense(y)
    # Test
    x_val = np.arange(48).reshape(4, 4, 3)
    y_val = tf.SparseTensorValue([[0, 0, 0], [2, 3, 0], [3, 1, 0]], [1, 2, 3], [4, 4, 1])
    res_dense_val, res_dense2_val = sess.run((res_dense, res_dense2),
                                             feed_dict={x: x_val, y: y_val})
    print(np.allclose(res_dense_val, res_dense2_val))
    # True