我们应该如何使用pad_sequences填充keras中的文本序列？

Question

我使用从网络教程中获得的知识和我自己的直觉，在keras中编写了一个sequence to sequence学习LSTM。我将示例文本转换为序列，然后使用keras中的pad_sequence函数进行填充。

from keras.preprocessing.text import Tokenizer,base_filter
from keras.preprocessing.sequence import pad_sequences

def shift(seq, n):
    n = n % len(seq)
    return seq[n:] + seq[:n]

txt="abcdefghijklmn"*100

tk = Tokenizer(nb_words=2000, filters=base_filter(), lower=True, split=" ")
tk.fit_on_texts(txt)
x = tk.texts_to_sequences(txt)
#shifing to left
y = shift(x,1)

#padding sequence
max_len = 100
max_features=len(tk.word_counts)
X = pad_sequences(x, maxlen=max_len)
Y = pad_sequences(y, maxlen=max_len)

仔细检查后，我发现我的填充序列看起来像这样

>>> X[0:6]
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7]], dtype=int32)
>>> X
array([[ 0,  0,  0, ...,  0,  0,  1],
       [ 0,  0,  0, ...,  0,  0,  3],
       [ 0,  0,  0, ...,  0,  0,  2],
       ..., 
       [ 0,  0,  0, ...,  0,  0, 13],
       [ 0,  0,  0, ...,  0,  0, 12],
       [ 0,  0,  0, ...,  0,  0, 14]], dtype=int32)

填充序列是否看起来像这样？除了数组中的最后一列，其余的都是零。我想在填充文本时我犯了一些错误，如果是这样，你能告诉我我在哪里犯了错误吗？

Answer 1

如果你想用char进行标记，你可以手动进行，这不是太复杂：

首先为你的角色建立一个词汇表：

txt="abcdefghijklmn"*100
vocab_char = {k: (v+1) for k, v in zip(set(txt), range(len(set(txt))))}
vocab_char['<PAD>'] = 0

这将为txt中的每个字符关联一个不同的数字。应为padding保留索引为0的字符。

具有反向词汇表将对解码输出有用。

rvocab = {v: k for k, v in vocab.items()}

一旦你有了这个，你可以先将你的文本分成序列，比如你想要有seq_len = 13长度的序列：

[[vocab_char[char] for char in txt[i:(i+seq_len)]] for i in range(0,len(txt),seq_len)]

您的输出将如下所示：

[[9, 12, 6, 10, 8, 7, 2, 1, 5, 13, 11, 4, 3], 
 [14, 9, 12, 6, 10, 8, 7, 2, 1, 5, 13, 11, 4],
 ...,
 [2, 1, 5, 13, 11, 4, 3, 14, 9, 12, 6, 10, 8], 
 [7, 2, 1, 5, 13, 11, 4, 3, 14]]

请注意，最后一个序列的长度不同，您可以将其丢弃或将序列填充到max_len = 13，它会为它添加0。

您可以通过将所有内容移动1来以相同的方式构建目标Y :-)

我希望这有帮助。

Answer 2

问题出在这一行：

tk = Tokenizer(nb_words=2000, filters=base_filter(), lower=True, split=" ")

当您设置此类分割（通过" "）时，由于数据的性质，您将获得由单个单词组成的每个序列。这就是你的填充序列只有一个非零元素的原因。要改变尝试：

txt="a b c d e f g h i j k l m n "*100