我一直在寻找能够在内存有限(约64MB)的java中读取和写入大文件(0.5 - 1 GB)的最快方法。文件中的每一行代表一条记录,所以我需要逐行获取它们。该文件是普通文本文件。
我尝试过BufferedReader和BufferedWriter,但它似乎不是最好的选择。读取和写入大小为0.5 GB的文件大约需要35秒,只读取没有处理的写入。我认为这里的瓶颈是写作,因为单独阅读大约需要10秒钟。
我尝试读取字节数组,但是在每个读取的数组中搜索行需要更多时间。
有什么建议吗? 感谢
答案 0 :(得分:15)
我怀疑你真正的问题是你的硬件有限,你所做的是软件不会有太大的区别。如果你有足够的内存和CPU,更高级的技巧可以帮助,但如果你只是等待你的硬盘驱动器,因为文件没有缓存,它将没有太大的区别。
BTW:10秒或500 MB /秒500 MB是硬盘的典型读取速度。
尝试运行以下内容,了解您的系统无法有效缓存文件。
public static void main(String... args) throws IOException {
for (int mb : new int[]{50, 100, 250, 500, 1000, 2000})
testFileSize(mb);
}
private static void testFileSize(int mb) throws IOException {
File file = File.createTempFile("test", ".txt");
file.deleteOnExit();
char[] chars = new char[1024];
Arrays.fill(chars, 'A');
String longLine = new String(chars);
long start1 = System.nanoTime();
PrintWriter pw = new PrintWriter(new FileWriter(file));
for (int i = 0; i < mb * 1024; i++)
pw.println(longLine);
pw.close();
long time1 = System.nanoTime() - start1;
System.out.printf("Took %.3f seconds to write to a %d MB, file rate: %.1f MB/s%n",
time1 / 1e9, file.length() >> 20, file.length() * 1000.0 / time1);
long start2 = System.nanoTime();
BufferedReader br = new BufferedReader(new FileReader(file));
for (String line; (line = br.readLine()) != null; ) {
}
br.close();
long time2 = System.nanoTime() - start2;
System.out.printf("Took %.3f seconds to read to a %d MB file, rate: %.1f MB/s%n",
time2 / 1e9, file.length() >> 20, file.length() * 1000.0 / time2);
file.delete();
}
在具有大量内存的Linux计算机上。
Took 0.395 seconds to write to a 50 MB, file rate: 133.0 MB/s
Took 0.375 seconds to read to a 50 MB file, rate: 140.0 MB/s
Took 0.669 seconds to write to a 100 MB, file rate: 156.9 MB/s
Took 0.569 seconds to read to a 100 MB file, rate: 184.6 MB/s
Took 1.585 seconds to write to a 250 MB, file rate: 165.5 MB/s
Took 1.274 seconds to read to a 250 MB file, rate: 206.0 MB/s
Took 2.513 seconds to write to a 500 MB, file rate: 208.8 MB/s
Took 2.332 seconds to read to a 500 MB file, rate: 225.1 MB/s
Took 5.094 seconds to write to a 1000 MB, file rate: 206.0 MB/s
Took 5.041 seconds to read to a 1000 MB file, rate: 208.2 MB/s
Took 11.509 seconds to write to a 2001 MB, file rate: 182.4 MB/s
Took 9.681 seconds to read to a 2001 MB file, rate: 216.8 MB/s
在具有大量内存的Windows计算机上。
Took 0.376 seconds to write to a 50 MB, file rate: 139.7 MB/s
Took 0.401 seconds to read to a 50 MB file, rate: 131.1 MB/s
Took 0.517 seconds to write to a 100 MB, file rate: 203.1 MB/s
Took 0.520 seconds to read to a 100 MB file, rate: 201.9 MB/s
Took 1.344 seconds to write to a 250 MB, file rate: 195.4 MB/s
Took 1.387 seconds to read to a 250 MB file, rate: 189.4 MB/s
Took 2.368 seconds to write to a 500 MB, file rate: 221.8 MB/s
Took 2.454 seconds to read to a 500 MB file, rate: 214.1 MB/s
Took 4.985 seconds to write to a 1001 MB, file rate: 210.7 MB/s
Took 5.132 seconds to read to a 1001 MB file, rate: 204.7 MB/s
Took 10.276 seconds to write to a 2003 MB, file rate: 204.5 MB/s
Took 9.964 seconds to read to a 2003 MB file, rate: 210.9 MB/s
答案 1 :(得分:9)
我要尝试的第一件事是增加BufferedReader和BufferedWriter的缓冲区大小。默认的缓冲区大小没有记录,但至少在Oracle VM中它们是8192个字符,这不会带来太多的性能优势。
如果您只需要复制文件(并且不需要实际访问数据),我会放弃Reader / Writer方法,并使用字节数组作为缓冲区直接使用InputStream和OutputStream:
FileInputStream fis = new FileInputStream("d:/test.txt");
FileOutputStream fos = new FileOutputStream("d:/test2.txt");
byte[] b = new byte[bufferSize];
int r;
while ((r=fis.read(b))>=0) {
fos.write(b, 0, r);
}
fis.close();
fos.close();
或实际使用NIO:
FileChannel in = new RandomAccessFile("d:/test.txt", "r").getChannel();
FileChannel out = new RandomAccessFile("d:/test2.txt", "rw").getChannel();
out.transferFrom(in, 0, Long.MAX_VALUE);
in.close();
out.close();
在对不同的复制方法进行基准测试时,我在每次运行基准测试之间的差异(持续时间)要大于不同实现之间的差异。 I / O缓存(在操作系统级别和硬盘缓存上)在这里发挥了重要作用,很难说什么更快。在我的硬件上,使用BufferedReader和BufferedWriter逐行复制1GB文本文件在某些运行中少于5秒,在其他运行中则少于30秒。
答案 2 :(得分:4)
在Java 7中,您可以使用Files.readAllLines()和Files.write()方法。这是一个例子:
List<String> readTextFile(String fileName) throws IOException {
Path path = Paths.get(fileName);
return Files.readAllLines(path, StandardCharsets.UTF_8);
}
void writeTextFile(List<String> strLines, String fileName) throws IOException {
Path path = Paths.get(fileName);
Files.write(path, strLines, StandardCharsets.UTF_8);
}
答案 3 :(得分:2)
我建议查看java.nio
包中的类。套接字的非阻塞IO可能更快:
http://docs.oracle.com/javase/6/docs/api/java/nio/package-summary.html
本文的基准测试表明这是真的:
http://vanillajava.blogspot.com/2010/07/java-nio-is-faster-than-java-io-for.html
答案 4 :(得分:0)
我写了一篇关于reading files in Java的许多方法的大量文章,并用1KB到1GB的样本文件相互测试它们,我发现以下3种方法读取1GB文件的速度最快: / p>
1)java.nio.file.Files.readAllBytes() - 用不到1秒的时间读取1 GB的测试文件。
import java.io.File;
import java.io.IOException;
import java.nio.file.Files;
public class ReadFile_Files_ReadAllBytes {
public static void main(String [] pArgs) throws IOException {
String fileName = "c:\\temp\\sample-10KB.txt";
File file = new File(fileName);
byte [] fileBytes = Files.readAllBytes(file.toPath());
char singleChar;
for(byte b : fileBytes) {
singleChar = (char) b;
System.out.print(singleChar);
}
}
}
2)java.nio.file.Files.lines() - 在1 GB的测试文件中读取大约需要3.5秒。
import java.io.File;
import java.io.IOException;
import java.nio.file.Files;
import java.util.stream.Stream;
public class ReadFile_Files_Lines {
public static void main(String[] pArgs) throws IOException {
String fileName = "c:\\temp\\sample-10KB.txt";
File file = new File(fileName);
try (Stream linesStream = Files.lines(file.toPath())) {
linesStream.forEach(line -> {
System.out.println(line);
});
}
}
}
3)java.io.BufferedReader - 花了大约4.5秒来读取1 GB的测试文件。
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
public class ReadFile_BufferedReader_ReadLine {
public static void main(String [] args) throws IOException {
String fileName = "c:\\temp\\sample-10KB.txt";
FileReader fileReader = new FileReader(fileName);
try (BufferedReader bufferedReader = new BufferedReader(fileReader)) {
String line;
while((line = bufferedReader.readLine()) != null) {
System.out.println(line);
}
}
}
}
答案 5 :(得分:0)
关于OutOfMemoryException
的全部内容都可以通过Scanner类迭代器有效地处理。它逐行读取文件,而不是批量读取。
以下代码解决了该问题:
try(FileInputStream inputStream =new FileInputStream("D:\\File\\test.txt");
Scanner sc= new Scanner(inputStream, "UTF-8")) {
while (sc.hasNextLine()) {
String line = sc.nextLine();
System.out.println(line);
}
} catch (IOException e) {
e.printStackTrace();
}