Question

tensorflow有tf.string_split函数，可以将密集张量分割为SparseTensor，但不提供相反的函数。

谁知道怎么做？感谢〜

例如： SparseTensor：

[["a", "b", "c"]
 ["d", "e"]
 ["f", "g", "h", "i"]]

将SparseTensor与分隔符“”连接到密集张量：

["a b c",
 "d e",
 "f g h i"]

tensorflow中SparseTensor格式：

# indices = tf.constant([[0, 0], [0, 1], [0, 2], [1, 0], [1, 1], [2, 0], [2, 1], [2, 2], [2, 3]], dtype=tf.int64)
# values = tf.constant(["a", "b", "c", "d", "e", "f", "g", "h", "i"], dtype=tf.string)
# dense_shape = tf.constant([3, 4], dtype=tf.int64)
# tf.SparseTensor(indices=indices, values=values, dense_shape=dense_shape)

更新问题：

join_words_list = []
slice_words_list = tf.sparse_split(sp_input=sparse_words, num_split=3, axis=0)
# slice_words_list = tf.sparse_split(sp_input=sparse_words, num_split=sparse_words.dense_shape[0], axis=0)
for slice_words in slice_words_list:
  slice_words = slice_words.values
  join_words = tf.reduce_join(slice_words, reduction_indices=0, separator=" ")
  join_words_list.append(join_words)
join_str = tf.stack(join_words_list)

现在，这会产生新问题，tensorflow, split tf.string SparseTensor to the list of dense Tensor in dim 0

Answer 1

这是一个sparse_string_join实现，可以实现您所需要的：

def sparse_string_join(input_sp):
  """Concats each row of SparseTensor `input_sp` and outputs them as a 1-D string tensor."""

  # convert the `SparseTensor` to a dense `Tensor`
  dense_input = tf.sparse_to_dense(input_sp.indices, input_sp.dense_shape, input_sp.values, default_value='')

  # concat each row of strings.
  strings = tf.reduce_join(dense_input, axis=1, separator=' ')

  # remove extra spaces.  
  return tf.string_strip(strings)

Answer 2

查看string_join

output = tf.string_join([input0, input1])

tensorflow，如何将tf.string SparseTensor加入到1-D密集张量中

2 个答案: