用于捕获缩写的正则表达式

Question

我正在尝试制作一个匹配字符串中的缩写及其完整形式的正则表达式。我有一个捕获某些情况的正则表达式，但在下面的示例中，它捕获的单词多于应有的单词。谁能帮我解决这个问题？

x = 'Confirmatory factor analysis (CFA)  is a special case of what is known as structural equation modelling (SEM).'

re.findall(r'\b([A-Za-z][a-z]+(?:\s[A-Za-z][a-z]+)+)\s+\(([A-Z][A-Z]*[A-Z]\b\.?)',x)

出来：

[('Confirmatory factor analysis', 'CFA'),
 ('special case of what is known as structural equation modeling', 'SEM')]

Answer 1

1. 只有一种方法可以知道（CFA）之前有多少单词构成所谓的完整形式：查看第 2 组中的 alpha 数量（分配给l），将第 1 组拆分为空格，根据第2组的长度取最后一个l单词，然后重新加入。

2. 您的正则表达式将接受 (CFA.) 但不接受 (C.F.A.)，因此对您的正则表达式稍作修改是为了在每个 alpha 之后允许一个可选的句点，并且您似乎试图说缩写必须包含两个或多个字母字符 - 有一种更简单的方式来表达。

在正则表达式中更改为第 2 组：

(                    # start of group 2
  (?:                # start of non-capturing group
     [A-Z]           # an alpha character
     \.?             # optionally followed by a period
  )                  # end of non-capturing group
  {2,}               # the non-capturing group is repeated 2 or more times
)                    # end of group 2

代码：

#!/usr/bin/env python3

import re

x = 'Confirmatory factor analysis (CFA)  is a special case of what is known as structural equation modelling (S.E.M.).'
results = []
split_regex = re.compile(r'\s+')
for m in re.finditer(r'\b([A-Za-z][a-z]*(?:\s[A-Za-z][a-z]*)+)\s+\(((?:[A-Z]\.?){2,})\)', x):
    abbreviation = m[2]
    l = sum(c.isalpha() for c in abbreviation)
    full_form = ' '.join(split_regex.split(m[1])[-l:])
    results.append([full_form, abbreviation])
print(results)

打印

[['Confirmatory factor analysis', 'CFA'], ['structural equation modelling', 'S.E.M.']]

Python Demo

Answer 2

试试这个——它通过查找用括号括起来的大写字符串来工作。然后我们验证前面的单词是否匹配 abbrv。

import re

string = 'Confirmatory factor analysis (CFA)  is a special case of what is known as structural equation modelling (SEM).'

abbrvs  = re.findall("\(([A-Z][A-Z]+)\)", string) #find potential abbrvs

words = re.split("\s|\.|,", string) 

validated_abbrvs = []
for abbrv in abbrvs:
    end = words.index(f"({abbrv})")
    start = end - len(abbrv) 
    full_name = words[start:end] #locate preceeding words
    if "".join([w[0].upper() for w in full_name]) == abbrv: #validate it matches abbrv
        validated_abbrvs.append((abbrv, " ".join(full_name)))

print(validated_abbrvs)

Answer 3

我使用了正则表达式并用 ( 或 ) 分割字符串。 然后在顺序索引中创建一个元组列表。

import re
x = 'Confirmatory factor analysis (CFA)  is a special case of what is known as structural equation modelling (SEM).'
lst = re.split('\(|\)', x)
lst = [(lst[i*2].strip(), lst[i*2+1].strip()) for i in range(0, len(lst)//2)]
final = []
for i in range(len(lst)):
    abbr = lst[i][1]
    text = ' '.join(lst[i][0].split(' ')[-len(abbr):])
    final.append((abbr, text)) 
final

结果：

 [('CFA', 'Confirmatory factor analysis'),
 ('SEM', 'structural equation modelling')]