如何在Python中使用spacy依赖树获得祖先的孩子

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)])])

其输出是：
[(u'Photos', u'NOUN', u'nsubj', [(u'are', u'VERB')]

现在我希望打印acomp以下子项：
[(u'are', u'VERB')]
这是以下项的祖先：
[(u'Photos', u'NOUN', u'nsubj')]

我怎样才能做到这一点？

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)

其输出（在Python3中）：

photo's are:
poor

不过，请查看Spacy 的关于依赖关系的页面。有很多事情需要考虑。您可以使用DisplaCy（此链接也是类似句子的示例，具有不同的依赖关系）。

我希望这至少可以帮助您指明正确的方向！