Java 在这些行中查找特定模式的正确正则表达式是什么？

所以我有一个大文件，其中包含了大量的天气数据。我必须将大文件中的每一行分配到相应的状态文件中。因此，总共将有50个新的状态文件，其中包含它们自己的数据

大文件包含约100万行记录，如下所示：

COOP:166657,'NEW IBERIA AIRPORT ACADIANA REGIONAL LA US',200001,177,553

虽然电台的名称可能不同，字数也可能不同

这是我正在使用的正则表达式：

Pattern p = Pattern.compile(".* ([A-Z][A-Z]) US.*"); 
Matcher m = p.matcher(line);

当我运行我的程序时，仍然存在无法找到模式的行实例

这是我的节目：

package climate;

import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.util.Arrays;
import java.util.Scanner;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

/**
 * This program will read in a large file containing many stations and states,
 * and output in order the stations to their corresponding state file.
 * 
 * Note: This take a long time depending on processor. It also appends data to
 * the files so you must remove all the state files in the current directory
 * before running for accuracy.
 * 
 * @author Marcus
 *
 */

public class ClimateCleanStates {

    public static void main(String[] args) throws IOException {

        Scanner in = new Scanner(System.in);
        System.out
                .println("Note: This program can take a long time depending on processor.");
        System.out
                .println("It is also not necessary to run as state files are in this directory.");
        System.out
                .println("But if you would like to see how it works, you may continue.");
        System.out.println("Please remove state files before running.");
        System.out.println("\nIs the States directory empty?");
        String answer = in.nextLine();

        if (answer.equals("N")) {
            System.exit(0);
            in.close();
        }
        System.out.println("Would you like to run the program?");
        String answer2 = in.nextLine();
        if (answer2.equals("N")) {
            System.exit(0);
            in.close();
        }

        String[] statesSpaced = new String[51];

        File statefile, dir, infile;

        // Create files for each states
        dir = new File("States");
        dir.mkdir();


        infile = new File("climatedata.csv");
        FileReader fr = new FileReader(infile);
        BufferedReader br = new BufferedReader(fr);

        String line;
        System.out.println();

        // Read in climatedata.csv
        // Probably need to implement ClimateRecord class
        final long start = System.currentTimeMillis();
        while ((line = br.readLine()) != null) {
            // Remove instances of -9999

            if (!line.contains("-9999")) {



                        Pattern p = Pattern.compile("^.* ([A-Z][A-Z]) US.*$"); 
                        Matcher m = p.matcher(line);
                        String stateFileName = null;

                        if(m.find()){
                            //System.out.println(m.group(1));
                            stateFileName = m.group(1);
                        } else {
                            System.out.println("Could not find abbreviation");
                        }

                        /*
                        stateFileName = "States/" + stateFileName + ".csv";
                        statefile = new File(stateFileName);

                        FileWriter stateWriter = new FileWriter(statefile, true);
                        stateWriter.write(line + "\n");
                        // Progress reporting
                        System.out.printf("Writing [%s] to file [%s]\n", line,
                                statefile);
                        stateWriter.flush();
                        stateWriter.close();
                        */





            }
        }
        System.out.println("Elapsed " + (System.currentTimeMillis() - start) + " ms");
        br.close();
        fr.close();
        in.close();

    }

}

---

取决于要提取的内容，但如果使用类型的模式

Pattern.compile("(.*):(.*),'(.*)',(.*),(.*),(.*)");
Matcher m = p.matcher(line);
if(m.find()) {
  // here you can use with i from 1 to 6
  m.group(i); 

  //and access the 6 tokens:
  //COOP
  //166657
  //NEW IBERIA AIRPORT ACADIANA REGIONAL LA US
  //200001
  //177
  //553
}

而不是

“*（[A-Z][A-Z]）美国。*”

如果某些州没有缩写，请尝试：

---

“（[a-z][a-z]）+US”

请特别注意

^

行的开头、非贪婪组

（.*）

、行的结尾

$

、

点球
和
多行

Pattern regex = Pattern.compile("^(.*?):(.*?),'(.*?)',(.*?),(.*?),(.*?)$", Pattern.DOTALL | Pattern.MULTILINE);


正则表达式演示：



Live JAVA示例：



正则表达式解释：

^(.*?):(.*?),'(.*?)',(.*?),(.*?),(.*?)$

Options: Case sensitive; Exact spacing; Dot matches line breaks; ^$ match at line breaks; Default line breaks

Assert position at the beginning of a line (at beginning of the string or after a line break character) (carriage return and line feed, next line, line separator, paragraph separator) «^»
Match the regex below and capture its match into backreference number 1 «(.*?)»
   Match any single character «.*?»
      Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the character “:” literally «:»
Match the regex below and capture its match into backreference number 2 «(.*?)»
   Match any single character «.*?»
      Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the character string “,'” literally «,'»
Match the regex below and capture its match into backreference number 3 «(.*?)»
   Match any single character «.*?»
      Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the character string “',” literally «',»
Match the regex below and capture its match into backreference number 4 «(.*?)»
   Match any single character «.*?»
      Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the character “,” literally «,»
Match the regex below and capture its match into backreference number 5 «(.*?)»
   Match any single character «.*?»
      Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the character “,” literally «,»
Match the regex below and capture its match into backreference number 6 «(.*?)»
   Match any single character «.*?»
      Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Assert position at the end of a line (at the end of the string or before a line break character) (carriage return and line feed, next line, line separator, paragraph separator) «$»

我认为您需要环顾函数，它们断言某些内容应该在匹配的表达式之前或之后，但不包括在结果中

(?<= )[A-Z][A-Z](?= US)

（？您可以通过以下方式验证它是否是美国各州的缩写：

\s(?:(A[KLRZ]|C[AOT]|D[CE]|FL|GA|HI|I[ADLN]|K[SY]|LA|M[ADEINOST]|N[CDEHJMVY]|O[HKR]|P[AR]|RI|S[CD]|T[NX]|UT|V[AIT]|W[AIVY])(?:\sUS'|'))

正如我所见，站点名称被单引号包围，为什么不搜索成对的站点名称？事实证明，有些行不包含状态缩写。我会根据站点名称输出文件，但必须按状态分组。您知道如果字符串为是两个字母而不是“我们”，还是只做正则表达式？或者有更快的方法来实现这一点？拆分可能会更快，但正则表达式引擎也非常快：为了区分两者的不同，请编写两个定时测试。请注意，您可以从命令行使用awk和sed等工具组合轻松筛选文件。