使用从文本文件中删除所有数字和字母数字字符

Question

我有 2 个文本文件：

File1 - 这个文件的格式是user_id tweet_id tweet_text

文件 1

60730027    6298443824  thank you echo park. you've changed A LOT, but as long as I'm getting paid to make you move, I'm still with it! 2009-12-03 02:54:10
60730027    6297282530  fat Albert Einstein goin in right now over here!!!  2009-12-03 01:35:22

文件2
这个文件的格式是genome_id name ascii_name

4045417 Southwest Indent    Southwest Indent
4045418 Southeast Point     Southeast Point     

下面是sn-p读取文件1的代码：

public void readfromFile() throws FileNotFoundException {
    Scanner inputStream;
    String source=null;
     FileInputStream file = new FileInputStream("file1.txt");   
        String regex = "/[a-zA-Z ]+/";
        Scanner fileScan = new Scanner(file); 

        while(fileScan.hasNextLine()){
            word = fileScan.nextLine();
            word = word.replaceAll(regex, "").toLowerCase();
            PrintWriter outputStreamName = new PrintWriter(new FileOutputStream("temp.txt"));
            outputStreamName.printf("%s",word);
}

我的意图首先是用空值替换 user_id、tweet_id、genome_id 中存在的数据。然后将大写值转换为小写。但是，现在每当此代码处理 file1 时，文本文件都不会发生任何变化。我也想知道发生了什么。当我将它输出到控制台时，我得到了输出。

预期输出：

thank you echo park youve changed a lot but as long as im getting paid to make you move im still with it

fat albert einstein goin in right now over here

Answer 1

根据预期输出，您想要替换除字母、点和单词之间的空格以外的所有内容。

[^a-zA-Z. ]+|(?<=\d)\s*(?=\d)|(?<=\D)\s*(?=\d)|(?<=\d)\s*(?=\D)

这里是online demo

或者尝试不使用Lookaround

[^a-zA-Z. ]+|\d\s+\d|\D\s+\d|\d\s+\D

这里\s匹配任何空白字符[\r\n\t\f ]

示例代码：

String regex = "[^a-zA-Z. ]+|(?<=\\d)\\s*(?=\\d)|(?<=\\D)\\s*(?=\\d)|(?<=\\d)\\s*(?=\\D)";
str.replaceAll(regex,"");

输出：

thank you echo park. youve changed A LOT but as long as Im getting paid to make you move Im still with it
fat Albert Einstein goin in right now over here

---

要从输出中排除'，请使用[^a-zA-Z.' ]+，否则I'm 和you've 将更改为Im 和youve。

更好使用[a-zA-Z']+ 仅获取所有单词。这里是demo

示例代码：

String str = "60730027    6297282530  fat Albert Einstein goin in right now over here!!!  2009-12-03 01:35:22 ";
Pattern p = Pattern.compile("[a-zA-Z']+");
Matcher m = p.matcher(str);
while (m.find()) {
    System.out.print(m.group()+" ");
}

输出：

fat Albert Einstein goin in right now over here 

---

注意：因此您正在检查下一行

变化：

source = inputStream.next();

收件人：

source = inputStream.nextLine();

Answer 2

public void readfromFile() throws Exception
{

    FileInputStream file = new FileInputStream("file1.txt");    

    StringBuilder builder = new StringBuilder();
    int ch;
    while((ch = file.read()) != -1){
        builder.append((char)ch);
    }

    System.out.println(builder.toString().replaceAll("[^a-zA-Z\\s]", ""));

}

扫描仪过滤空字符串。

前任

Scanner scanner = new Scanner("60730027    6298443824  thank");
while(scanner.hasNext())    //Read from file till the last line of the file.
{
    System.out.print(scanner.next());
}

输出

607300276298443824谢谢

所以我们不能使用扫描仪。

Answer 3

试试这个

s = s.replaceAll("\\d+\\s+\\d+\\s+", "").replaceAll(" +\\S+ \\S+$", "");