基于 Spacy 令牌的匹配，令牌之间的令牌数为 'n'

Question

我正在使用 spacy 来匹配某些文本（意大利语）中的特定表达式。我的文本可以以多种形式出现，我正在尝试了解编写一般规则的最佳方式是什么。我有以下 4 个案例，我想写一个适用于所有案例的通用模式。类似于：

# case 1
text = 'Superfici principali e secondarie: 90 mq'
# case 2
# text = 'Superfici principali e secondarie di 90 mq'
# case 3
# text = 'Superfici principali e secondarie circa 90 mq'
# case 4
# text = 'Superfici principali e secondarie di circa 90 mq'

nlp = spacy.load('it_core_news_sm')
doc = nlp(text)

matcher = Matcher(nlp.vocab) 

pattern = [{"LOWER": "superfici"}, {"LOWER": "principali"}, {"LOWER": "e"}, {"LOWER": "secondarie"},  << "some token here that allows max 3 tokens or a IS_PUNCT or nothing at all" >>, {"IS_DIGIT": True}, {"LOWER": "mq"}]

matcher.add("Superficie", None, pattern)

matches = matcher(doc)
for match_id, start, end in matches:
    string_id = nlp.vocab.strings[match_id]  # Get string representation
    span = doc[start:end]  # The matched span
    print(match_id, string_id, start, end, span.text)

Answer 1

您可以添加一个 {"IS_PUNCT": True, "OP": "?"} 可选标记，然后添加三个可选 IS_ALPHA 标记：

pattern = [
            {"LOWER": "superfici"}, 
            {"LOWER": "principali"},
            {"LOWER": "e"},
            {"LOWER": "secondarie"},
            {"IS_PUNCT": True, "OP": "?"},
            {"IS_ALPHA": True, "OP": "?"},
            {"IS_ALPHA": True, "OP": "?"},
            {"IS_ALPHA": True, "OP": "?"},
            {"IS_DIGIT": True},
            {"LOWER": "mq"}
          ]

"OP" : "?"表示令牌可以重复 1 次或 0 次，即它只能出现一次或消失。

基于 Spacy 令牌的匹配，令牌之间的令牌数为 'n'

Spacy token-based matching with 'n' number of tokens between tokens

python

nlp

spacy