java - 从文本中提取信息

Question

我有以下文字:

Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book.              

Name                                 Group                       12345678        
ALEX A ALEX                                                                   
ID#                                  PUBLIC NETWORK                  
XYZ123456789                                                                  


Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book.

我想提取文本中 ID# 关键字下的 ID 值。

问题是在不同的文本文件中 ID 可以位于不同的位置，例如在另一个文本的中间，如下所示:

Lorem Ipsum is simply dummy text of                                          ID#             the printing and typesetting industry. Lorem Ipsum has been the industry's          
standard dummy text ever since the 1500s, when an unknown printer took a     XYZ123456789    galley of type and scrambled it to make a type specimen book.       

此外，ID# 和值之间可以有额外的行:

Lorem Ipsum is simply dummy text of                                          ID#             the printing and typesetting industry. Lorem Ipsum has been the industry's      
printing and typesetting industry. Lorem Ipsum has been the                                  printing and typesetting industry. Lorem Ipsum has been the 
standard dummy text ever since the 1500s, when an unknown printer took a     XYZ123456789    galley of type and scrambled it to make a type specimen book.

您能否展示一种如何提取上述 ID# 值的方法？是否可以在此处应用任何标准技术来提取此信息？例如 RegEx 或 RegEx 之上的某种方法。这里可以应用NLP吗？

Answer 1

ID 值似乎没有明确的格式，因此单行正则表达式无济于事，因为这里几乎没有任何正则。

您必须使用两个正则表达式才能获得预期的输出。第一个是:

(?m)^(.*)ID#.*([\s\S]*)

它尝试查找ID#单独成行。它捕获两 block 字符串。第一个 block 是从该行开头到 ID# 的所有内容。那么 ID# 行之后出现的所有内容驻留。

然后我们计算第一个捕获组的长度。它为我们提供了列号，我们应该在下一行中开始搜索 ID:

m.group(1).length();

然后我们构建使用此长度的第二个正则表达式:

(?m)^.{X}(?<!\S)\h{0,3}(\S+)

分割:

• (?m)启用多行模式
• ^匹配行首
• .{X}匹配前 X 个字符(X 为 m.group(1).length() )
• (?<!\S)检查当前位置是否在空格字符之前
• \h{0,3}匹配水平空格，可选最多 3 个字符(如果值向右移动)
• (\S+)捕获以下非空白字符

然后我们在之前的正则表达式的第二个捕获组上运行这个正则表达式:

Matcher m = Pattern.compile("(?m)^(.*)ID#.*([\\s\\S]*)").matcher(string);                  
if (m.find()) {
    Matcher m1 = Pattern.compile("(?m)^.{" + m.group(1).length() + "}(?<!\\S)\\h{0,3}(\\S+)").matcher(m.group(2));
    if (m1.find())
        System.out.println(m1.group(1));
}

Live demo