我需要使用Java逐行读取大约5-6 GB的大型文本文件。

我怎样才能快速做到呢?


当前回答

下面是一个完整的错误处理示例,并支持java 7之前的字符集规范。在Java 7中,您可以使用try-with-resources语法,这使得代码更简洁。

如果你只想要默认字符集,你可以跳过InputStream,使用FileReader。

InputStream ins = null; // raw byte-stream
Reader r = null; // cooked reader
BufferedReader br = null; // buffered for readLine()
try {
    String s;
    if (true) {
        String data = "#foobar\t1234\n#xyz\t5678\none\ttwo\n";
        ins = new ByteArrayInputStream(data.getBytes());
    } else {
        ins = new FileInputStream("textfile.txt");
    }
    r = new InputStreamReader(ins, "UTF-8"); // leave charset out for default
    br = new BufferedReader(r);
    while ((s = br.readLine()) != null) {
        System.out.println(s);
    }
}
catch (Exception e)
{
    System.err.println(e.getMessage()); // handle exception
}
finally {
    if (br != null) { try { br.close(); } catch(Throwable t) { /* ensure close happens */ } }
    if (r != null) { try { r.close(); } catch(Throwable t) { /* ensure close happens */ } }
    if (ins != null) { try { ins.close(); } catch(Throwable t) { /* ensure close happens */ } }
}

下面是Groovy版本,有完整的错误处理:

File f = new File("textfile.txt");
f.withReader("UTF-8") { br ->
    br.eachLine { line ->
        println line;
    }
}

其他回答

要做到这一点,

例如:

如果在当前目录中有dataFile.txt

import java.io.*;
import java.util.Scanner;
import java.io.FileNotFoundException;

public class readByLine
{
    public readByLine() throws FileNotFoundException
    {
        Scanner linReader = new Scanner(new File("dataFile.txt"));

        while (linReader.hasNext())
        {
            String line = linReader.nextLine();
            System.out.println(line);
        }
        linReader.close();

    }

    public static void main(String args[])  throws FileNotFoundException
    {
        new readByLine();
    }
}

输出如下:

下面是一个完整的错误处理示例,并支持java 7之前的字符集规范。在Java 7中,您可以使用try-with-resources语法,这使得代码更简洁。

如果你只想要默认字符集,你可以跳过InputStream,使用FileReader。

InputStream ins = null; // raw byte-stream
Reader r = null; // cooked reader
BufferedReader br = null; // buffered for readLine()
try {
    String s;
    if (true) {
        String data = "#foobar\t1234\n#xyz\t5678\none\ttwo\n";
        ins = new ByteArrayInputStream(data.getBytes());
    } else {
        ins = new FileInputStream("textfile.txt");
    }
    r = new InputStreamReader(ins, "UTF-8"); // leave charset out for default
    br = new BufferedReader(r);
    while ((s = br.readLine()) != null) {
        System.out.println(s);
    }
}
catch (Exception e)
{
    System.err.println(e.getMessage()); // handle exception
}
finally {
    if (br != null) { try { br.close(); } catch(Throwable t) { /* ensure close happens */ } }
    if (r != null) { try { r.close(); } catch(Throwable t) { /* ensure close happens */ } }
    if (ins != null) { try { ins.close(); } catch(Throwable t) { /* ensure close happens */ } }
}

下面是Groovy版本,有完整的错误处理:

File f = new File("textfile.txt");
f.withReader("UTF-8") { br ->
    br.eachLine { line ->
        println line;
    }
}

一种常见的模式是使用

try (BufferedReader br = new BufferedReader(new FileReader(file))) {
    String line;
    while ((line = br.readLine()) != null) {
       // process the line.
    }
}

如果假设没有字符编码,则可以更快地读取数据。例如,ASCII-7,但它不会有太大的区别。很有可能您对数据的处理将花费更长的时间。

EDIT:一种不太常用的模式,可以避免行泄漏的范围。

try(BufferedReader br = new BufferedReader(new FileReader(file))) {
    for(String line; (line = br.readLine()) != null; ) {
        // process the line.
    }
    // line is not visible here.
}

更新:在Java 8中你可以这样做

try (Stream<String> stream = Files.lines(Paths.get(fileName))) {
        stream.forEach(System.out::println);
}

注意:你必须将Stream放在try-with-resource块中,以确保对其调用#close方法,否则底层文件句柄永远不会关闭,直到GC在很久之后才关闭。

我的阅读习惯通常很简单:

void readResource(InputStream source) throws IOException {
    BufferedReader stream = null;
    try {
        stream = new BufferedReader(new InputStreamReader(source));
        while (true) {
            String line = stream.readLine();
            if(line == null) {
                break;
            }
            //process line
            System.out.println(line)
        }
    } finally {
        closeQuiet(stream);
    }
}

static void closeQuiet(Closeable closeable) {
    if (closeable != null) {
        try {
            closeable.close();
        } catch (IOException ignore) {
        }
    }
}

FileReader不会让你指定编码,如果你需要指定它,使用inputstreamreader代替:

try {
    BufferedReader br = new BufferedReader(new InputStreamReader(new FileInputStream(filePath), "Cp1252"));         

    String line;
    while ((line = br.readLine()) != null) {
        // process the line.
    }
    br.close();

} catch (IOException e) {
    e.printStackTrace();
}

如果从Windows导入该文件,它可能具有ANSI编码(Cp1252),因此必须指定编码。