2

[None, None]我有一个形状为 type的张量占位符string。例如，它看起来像这样

[["Hello", "World"], ["Amercian", "people"]].

现在我想将这个 2D 张量转换为 3D 张量，它现在基本上将每个单词剥离为其组成字符。所以输出看起来像 [[["H", "e", "l", "l", "o"], ["W", "o", "r", "l", "d"]], [["A", "m", "e", "r", "i", "c", "a", "n"], ["p", "e", "o", "p", "l", "e"]]].

由于每个单词都有不同数量的字符，因此新张量应该用空格填充小单词。有没有办法在张量流中解决这个问题？

4

1 回答 1

1

这运行

import tensorflow as tf
import tensorflow_transform as tft

input_data = tf.placeholder(shape=[None, None], dtype=tf.string, name="words")
words_flatten = tf.reshape(words, [tf.shape(words)[0] * tf.shape(words)[1]])
words_split = tf.string_split(words_flatten, delimiter="")
ngrams = tft.ngrams(words_split, ngram_range=(1,3), separator="")
tokens= tf.sparse_reset_shape(tf.sparse_fill_empty_rows(ngrams, "")[0])
tokens_dense = tf.reshape(
            tf.sparse_to_dense(tokens.indices, tokens.dense_shape, tokens.values, default_value=""),
            [tf.shape(words)[0], tf.shape(words)[1], -1]
        )

tokens_dense是所需的输出。

于 2019-01-22T07:18:39.737 回答