在Java中读取相对较大的字节文件的最快方法 [英] Fastest way of reading relatively huge byte-files in Java

问题描述

用Java的I / O方法读取相对较大的文件可能是最快的方法？我目前的解决方案使用 BufferedInputStream 保存到分配有1024字节的字节数组中。每个缓冲区都保存在 ArrayList 中以供以后使用。整个过程是通过一个单独的线程（callable-interface）调用的。

虽然速度不是很快。

  ArrayList< byte []> outputArr = new ArrayList< byte []>（）; 
 try {
 BufferedInputStream reader = new BufferedInputStream（new FileInputStream（dir + filename））; 
 
 byte [] buffer = new byte [LIMIT]; // == 1024 
 int i = 0; 
 while（reader.available（）！= 0）{
 reader.read（buffer）; 
 i ++; 
 if（i <= LIMIT）{
 outputArr.add（buffer）; 
 i = 0; 
 buffer = null; 
 buffer =新字节[LIMIT]; 
} 
 else else; 
 
 
 System.out.println（FileReader-Elements：+ outputArr.size（）+w。+ buffer.length +byte each。）; 
  

解决方案

我会使用一个足够快的内存映射文件（b）在$ {

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $>
MappedByteBuffer buffer = channel.map（FileChannel.MapMode.READ_ONLY，0，channel.size（））;

//完成后
channel.close（）;

假定文件小于2 GB并且需要10毫秒或更少。 p>

what's the probably fastest way of reading relatively huge files with Java's I/O-methods? My current solution uses the BufferedInputStream saving to an byte-array with 1024 bytes allocated to it. Each buffer is than saved in an ArrayList for later use. The whole process is called via a separate thread (callable-interface).

Not very fast though.

    ArrayList<byte[]> outputArr = new ArrayList<byte[]>();      
    try {
        BufferedInputStream reader = new BufferedInputStream(new FileInputStream (dir+filename));

        byte[] buffer = new byte[LIMIT]; // == 1024 
            int i = 0;
            while (reader.available() != 0) {
                reader.read(buffer);
                i++;
                if (i <= LIMIT){
                    outputArr.add(buffer);
                    i = 0;
                    buffer = null;
                    buffer = new byte[LIMIT];
                }
                else continue;              
            }

         System.out.println("FileReader-Elements: "+outputArr.size()+" w. "+buffer.length+" byte each.");   

解决方案

I would use a memory mapped file which is fast enough to do in the same thread.

final FileChannel channel = new FileInputStream(fileName).getChannel();
MappedByteBuffer buffer = channel.map(FileChannel.MapMode.READ_ONLY, 0, channel.size());

// when finished
channel.close();

This assumes the file is smaller than 2 GB and will take 10 milli-seconds or less.

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