在这些行中查找特定模式的正确正则表达式是什么?

What is the correct regular expression for finding specific pattern in these lines?

所以我有一个包含一堆天气数据的大文件。我必须将大文件中的每一行分配到其相应的状态文件中。因此,总共将有 50 个具有自己数据的新状态文件。

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会有所不同并且有不同的字数。

这是我使用的正则表达式:

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。

这是我的程序:

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();

    }

}

取决于你想提取的内容,但如果你使用类型

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]) US.*"

如果某些州没有缩写,也许可以尝试:

" ([a-z][A-Z])+ US'"

特别注意行首^,非贪心组(.*?),行尾$DOTALLMULTILINE

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

正则表达式演示:

https://regex101.com/r/bX0rS3/1


现场JAVA示例:

http://ideone.com/uAUaJT


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) «$»

我认为您需要查看函数,它们断言某些内容应该位于您正在匹配的表达式之前或之后,但不包含在结果中。

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

(?<= ) 必须是 space 之前的

[A-Z][A-Z]正好两个大写字母

(?= US) 必须是 space 和

之后的字母 US

环顾四周可能会更稳健:例如,(?= US) 可能是 (?= US',)。

您可以通过以下方式验证它是美国各州的缩写:

\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'|'))

Demo