Question

我有一组已经匹配的关键字。这是一种医学背景，因此至少在我尝试进行的分析中，我已经提出了等效的方案：

我有一辆带镀铬1000英寸轮辋的汽车。

假设我想返回关键字 rims 的所有子单词/词组作为短语，其中 rims 已被SpaCy标记为实体。

>

在python中，这就是我正在做的事情：

test_phrases = nlp("""I have a car with chrome 100-inch rims.""")
print(test_phrases.cats)
for t in test_phrases:
    print('Token: {} || POS: {} || DEP: {} CHILDREN: {} || ent_type: {}'.format(t,t.pos_,t.dep_,[c for c in t.children],t.ent_type_))

Token: I || POS: PRON || DEP: nsubj CHILDREN: [] || ent_type: 
Token: have || POS: VERB || DEP: ROOT CHILDREN: [I, car, .] || 
ent_type: 
Token: a || POS: DET || DEP: det CHILDREN: [] || ent_type: 
Token: car || POS: NOUN || DEP: dobj CHILDREN: [a, with] || ent_type: 
Token: with || POS: ADP || DEP: prep CHILDREN: [rims] || ent_type: 
Token: chrome || POS: ADJ || DEP: amod CHILDREN: [] || ent_type: 
Token: 100-inch || POS: NOUN || DEP: compound CHILDREN: [] || ent_type: 
Token: rims || POS: NOUN || DEP: pobj CHILDREN: [chrome, 100-inch] || 
ent_type: 
Token: . || POS: PUNCT || DEP: punct CHILDREN: [] || ent_type: CARPART

所以，我要使用的东西是这样的：

test_matcher = Matcher(nlp.vocab)

test_phrase = ['']
patterns = [[{'ENT':'CARPART',????}] for kp in test_phrase]
test_matcher.add('CARPHRASE', None, *patterns)

调用 test_doc 上的 test_matcher 使其返回：

chrome 100-inch rims

Answer 1

我认为我找到了一个令人满意的解决方案，该解决方案在创建Spacy类对象时将起作用。您可以对其进行测试，以确保它可以与您的解决方案一起使用，然后在Spacy管道中添加到类似this的地方：

from spacy.matcher import Matcher

keyword_list = ['rims']
patterns = [[{'LOWER':kw}] for kw in keyword_list]

test_matcher.add('TESTPHRASE',None, *patterns)


 def add_children_matches(doc,keyword_matcher):
     '''Add children to match on original single-token keyword.'''
    matches = keyword_matcher(doc)
    for match_id, start, end in matches:
        tokens = doc[start:end]
        print('keyword:',tokens)
        # Since we are getting children for keyword, there should only be one token
        if len(tokens) != 1:
            print('Skipping {}. Too many tokens to match.'.format(tokens))
            continue
        keyword_token = tokens[0]
        sorted_children = sorted([c.i for c in keyword_token.children] + [keyword_token.i],reverse=False)
        print('keyphrase:',doc[min(sorted_children):max(sorted_children)+1])



doc = nlp("""I have a car with chrome 1000-inch rims.""")
add_children_matches(doc,test_matcher)

这给出了：

keyword: rims
keyphrase: chrome 1000-inch rims

编辑：要完全回答我自己的问题，您必须使用类似以下内容的

：

 def add_children_matches(doc,keyword_matcher):
     '''Add children to match on original single-token keyword.'''
    matches = keyword_matcher(doc)
    spans = []
    for match_id, start, end in matches:
        tokens = doc[start:end]
        print('keyword:',tokens)
        # Since we are getting children for keyword, there should only be one token
        if len(tokens) != 1:
            print('Skipping {}. Too many tokens to match.'.format(tokens))
            continue
        keyword_token = tokens[0]
        sorted_children = sorted([c.i for c in keyword_token.children] + [keyword_token.i],reverse=False)
        print('keyphrase:',doc[min(sorted_children):max(sorted_children)+1])


    start_char = doc[min(sorted_children):max(sorted_children)+1].start_char
    end_char = doc[min(sorted_children):max(sorted_children)+1].end_char

    span = doc.char_span(start_char, end_char,label='CARPHRASE')
    if span != None:
        spans.append(span)

    return doc

关键字及其子项匹配项的匹配器

1 个答案: