如何在 Spacy phrasmatcher 中获取短语计数

Question

我正在尝试 spaCy 的 PhraseMatcher。我使用了网站中给出的示例的改编版，如下所示。

color_patterns = [nlp(text) for text in ('red', 'green', 'yellow')]
product_patterns = [nlp(text) for text in ('boots', 'coats', 'bag')]
material_patterns = [nlp(text) for text in ('bat', 'yellow ball')]

matcher = PhraseMatcher(nlp.vocab)
matcher.add('COLOR', None, *color_patterns)
matcher.add('PRODUCT', None, *product_patterns)
matcher.add('MATERIAL', None, *material_patterns)

doc = nlp("yellow ball yellow lines")
matches = matcher(doc)
for match_id, start, end in matches:
    rule_id = nlp.vocab.strings[match_id]  # get the unicode ID, i.e. 'COLOR'
    span = doc[start : end]  # get the matched slice of the doc
    print(rule_id, span.text)

输出为

COLOR yellow
MATERIAL ball

我的问题是如何计算短语的数量，使我的输出看起来像指示黄色出现两次而球只出现一次。

COLOR Yellow (2)
MATERIAL ball (1)

Answer 1

是这样的吗？

from collections import Counter
from spacy.matcher import PhraseMatcher
color_patterns = [nlp(text) for text in ('red', 'green', 'yellow')]
product_patterns = [nlp(text) for text in ('boots', 'coats', 'bag')]
material_patterns = [nlp(text) for text in ('bat', 'yellow ball')]

matcher = PhraseMatcher(nlp.vocab)
matcher.add('COLOR', None, *color_patterns)
matcher.add('PRODUCT', None, *product_patterns)
matcher.add('MATERIAL', None, *material_patterns)
d = []
doc = nlp("yellow ball yellow lines")
matches = matcher(doc)
for match_id, start, end in matches:
    rule_id = nlp.vocab.strings[match_id]  # get the unicode ID, i.e. 'COLOR'
    span = doc[start : end]  # get the matched slice of the doc
    d.append((rule_id, span.text))
print("\n".join(f'{i[0]} {i[1]} ({j})' for i,j in Counter(d).items()))

输出：

COLOR yellow (2)
MATERIAL yellow ball (1)

如何在 Spacy phrasmatcher 中获取短语计数

How to get phrase count in Spacy phrasematcher

nlp

python-3.x

spacy