如何在 python 中使用 spacy 依赖树获得祖先的 children

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)])])

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

现在我想打印 acomp children of:
[(u'are', u'VERB')]
这是以下的祖先：
[(u'Photos', u'NOUN', u'nsubj')]

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)

哪个输出（在 Python3 中）：

photo's are:
poor

但是，请看一下Spacy's page on dependencies. There are a lot to consider. You could play around with DisplaCy（这个link也是一个相似句子的例子，它有不同的依存关系）。

我希望这至少能帮助您指明正确的方向！

如何在 python 中使用 spacy 依赖树获得祖先的 children

How can I get children of an ancestor using spacy dependency tree in python

python

nltk

spacy