如何使用conllu库创建令牌列表？

Question

我试图使用conllu库创建一个CoNLL-U文件，作为我正在进行的通用依赖项标记项目的一部分。

sentence = [['The', 'the', 'DET', ... ], ['big', big', 'ADJ', ... ], ['dog', 'dog', 'NOUN', ...], ...]

from collections import OrderedDict

def compile_sent(sent):
    sent_list = list()
    for i, tok_data in enumerate(sent):
        tok_id = i + 1
        tok = tok_data[0]
        lemma = tok_data[1]
        pos = tok_data[2]
        feats = tok_data[3]
        compiled_tok = OrderedDict({'id': tok_id, 'form': tok, 'lemma': lemma, 'upostag': pos, 'xpostag': None, 'feats': feats, 'head': None, 'deprel': None, 'deps': None, 'misc': None})
        sent_list.append(compiled_tok)
    sent_list = sent_list.serialize()
    return sent_list

print(compile_sent(sentence))

Traceback (most recent call last):
  File "/Users/me/PycharmProjects/UDParser/Rough_Work.py", line 103, in <module>
    print(compile_sent(sentence))
  File "/Users/me/PycharmProjects/UDParser/Rough_Work.py", line 99, in compile_sent
    sent_list = sent_list.serialize()
AttributeError: 'list' object has no attribute 'serialize'

data = """
# text = The big dog
1   The     the    DET    _    Definite=Def|PronType=Art   _   _   _   _
2   big     big    ADJ    _    Degree=Pos                  _   _   _   _
3   dog     dog    NOUN   _    Number=Sing                 _   _   _   _

"""

sentences = data.parse()
sentence = sentences[0]
print(sentence)

TokenList<The, quick, brown, fox, jumps, over, the, lazy, dog, .>

对此类型的列表运行serialize()方法将使其返回为CoNLL-U格式的字符串，如上面示例中的data。但是，当您试图在正常的python列表上运行它时，它就会中断。

我如何创建这样的tokenlist而不是普通的python列表对象？

Answer 1

将sent_list从普通列表更改为标记列表。

from conllu import TokenList
from collections import OrderedDict

def compile_sent(sent):
    sent_list = TokenList()
    # ... etc ...

您可以通过在REPL中使用help（TokenList)查看TokenList上的函数。