Question

代码如下：

    import spacy
    from nltk import Tree
    en_nlp = spacy.load('en')
    parsed = en_nlp(u"Photos under low lighting are poor, both front and back cameras.")
    print(u'sentence:{0}'.format(parsed.text))
    try2 = []
    print(u'parsed_sentence_children::{0}'.format([(x.text,x.pos_,x.dep_,[(x.text,x.dep_) for x in list(x.children)]) for x in parsed]))
    print("\n\n")
    for x in parsed:
        if x.pos_=="NOUN" and x.dep_=="nsubj":
            print(u'Noun and noun subject:{0}'.format(try2 =[(x.text,x.pos_,x.dep_,[(x.text,x.pos_)for x in list(x.ancestors)])])

输出到：

[（你＆＃39;照片＆＃39;，你＆＃39; NOUN＆＃39;，你＆＃39; nsubj＆＃39;，[（你＆＃39;，＆＃39; VERB＆＃39; ）]

现在我希望打印

的acomp个孩子

[（你＆＃39;，你＆＃39; VERB＆＃39;）]

是

的祖先

[（你＆＃39;照片＆＃39;，你＆＃39; NOUN＆＃39;，你＆＃39; nsubj＆＃39;）]

我该怎么做？

Answer 1

您可以遍历令牌：

import spacy

nlp = spacy.load('en')

text = 'Photos under low lighting are poor, both front and back cameras.'

for token in nlp(text):
    if token.dep_ == 'nsubj': # Or other forms of subjects / objects
        print(token.lemma_+"'s are:")
        for a in token.ancestors:
            if a.text == 'are': # Or however you determine your selection
                for atok in a.children:
                    if atok.dep_ == 'acomp': # Note, you should look for more than just acomp
                        print(atok.text)

哪些输出（在Python3中）：

photo's are:
poor

但是，请查看Spacy's page on dependencies。有很多事要考虑。你可以玩DisplaCy（这个链接也是一个类似句子的例子，它有不同的依赖关系）。

我希望这至少有助于让你指出正确的方向！

如何在python中使用spacy依赖树获取祖先的子代

1 个答案: