java加速文件读取

Question

我有一个1.7G文件，格式如下：

String Long String Long String Long String Long ... etc

本质上，String是一个键，Long是hashmap中的一个值，我感兴趣的是在运行应用程序中的任何其他内容之前进行初始化

我目前的代码是：

  RandomAccessFile raf=new RandomAccessFile("/home/map.dat","r");
                raf.seek(0);
                while(raf.getFilePointer()!=raf.length()){
                        String name=raf.readUTF();
                        long offset=raf.readLong();
                        map.put(name,offset);
                }

这需要大约12分钟来完成，我相信有更好的方法来完成这一点，所以我将感谢任何帮助或指针

谢谢

---

是否按照EJP建议更新

谢谢你的建议，我希望这就是你的意思。如果这是错误的，请纠正我

DataInputStream dis=null;
    try{
     dis=new DataInputStream(new BufferedInputStream(new FileInputStream("/home/map.dat")));
     while(true){
       String name=dis.readUTF();
       long offset=dis.readLong();
       map.put(name, offset);
     }
    }catch (EOFException eofe){
      try{
        dis.close();
      }catch (IOException ioe){
        ioe.printStackTrace();
      }
    }

Answer 1

我将构造该文件，以便在适当的位置使用它。i、 e.不以这种方式加载。由于您有可变长度的记录，您可以构造每个记录位置的数组，然后按顺序放置键，以便对数据执行二进制搜索。（或者您可以使用自定义哈希表）然后可以使用方法包装此文件，该方法隐藏数据实际上存储在文件中，而不是转换为数据对象

如果您执行所有这些操作，“加载”阶段将变得多余，您将不需要创建这么多对象

---

这是一个很长的例子，但希望能说明什么是可能的

import vanilla.java.chronicle.Chronicle;
import vanilla.java.chronicle.Excerpt;
import vanilla.java.chronicle.impl.IndexedChronicle;
import vanilla.java.chronicle.tools.ChronicleTest;

import java.io.IOException;
import java.util.*;

public class Main {
    static final String TMP = System.getProperty("java.io.tmpdir");

    public static void main(String... args) throws IOException {
        String baseName = TMP + "/test";
        String[] keys = generateAndSave(baseName, 100 * 1000 * 1000);

        long start = System.nanoTime();
        SavedSortedMap map = new SavedSortedMap(baseName);
        for (int i = 0; i < keys.length / 100; i++) {
            long l = map.lookup(keys[i]);
//            System.out.println(keys[i] + ": " + l);
        }
        map.close();
        long time = System.nanoTime() - start;

        System.out.printf("Load of %,d records and lookup of %,d keys took %.3f seconds%n",
                keys.length, keys.length / 100, time / 1e9);
    }

    static SortedMap<String, Long> generateMap(int keys) {
        SortedMap<String, Long> ret = new TreeMap<>();
        while (ret.size() < keys) {
            long n = ret.size();
            String key = Long.toString(n);
            while (key.length() < 9)
                key = '0' + key;
            ret.put(key, n);
        }
        return ret;
    }

    static void saveData(SortedMap<String, Long> map, String baseName) throws IOException {
        Chronicle chronicle = new IndexedChronicle(baseName);
        Excerpt excerpt = chronicle.createExcerpt();
        for (Map.Entry<String, Long> entry : map.entrySet()) {
            excerpt.startExcerpt(2 + entry.getKey().length() + 8);
            excerpt.writeUTF(entry.getKey());
            excerpt.writeLong(entry.getValue());
            excerpt.finish();
        }
        chronicle.close();
    }

    static class SavedSortedMap {
        final Chronicle chronicle;
        final Excerpt excerpt;
        final String midKey;
        final long size;

        SavedSortedMap(String baseName) throws IOException {
            chronicle = new IndexedChronicle(baseName);
            excerpt = chronicle.createExcerpt();
            size = chronicle.size();
            excerpt.index(size / 2);
            midKey = excerpt.readUTF();
        }

        // find exact match or take the value after.
        public long lookup(CharSequence key) {
            if (compareTo(key, midKey) < 0)
                return lookup0(0, size / 2, key);
            return lookup0(size / 2, size, key);
        }

        private final StringBuilder tmp = new StringBuilder();

        private long lookup0(long from, long to, CharSequence key) {
            long mid = (from + to) >>> 1;
            excerpt.index(mid);
            tmp.setLength(0);
            excerpt.readUTF(tmp);
            if (to - from <= 1)
                return excerpt.readLong();
            int cmp = compareTo(key, tmp);
            if (cmp < 0)
                return lookup0(from, mid, key);
            if (cmp > 0)
                return lookup0(mid, to, key);
            return excerpt.readLong();
        }

        public static int compareTo(CharSequence a, CharSequence b) {
            int lim = Math.min(a.length(), b.length());
            for (int k = 0; k < lim; k++) {
                char c1 = a.charAt(k);
                char c2 = b.charAt(k);
                if (c1 != c2)
                    return c1 - c2;
            }
            return a.length() - b.length();
        }

        public void close() {
            chronicle.close();
        }
    }

    private static String[] generateAndSave(String baseName, int keyCount) throws IOException {
        SortedMap<String, Long> map = generateMap(keyCount);
        saveData(map, baseName);
        ChronicleTest.deleteOnExit(baseName);

        String[] keys = map.keySet().toArray(new String[map.size()]);
        Collections.shuffle(Arrays.asList(keys));
        return keys;
    }
}

生成2GB的原始数据并执行一百万次查找。它的编写方式使得加载和查找使用的堆很少。（<；<；1MB）

ls -l /tmp/test*
-rw-rw   1 peter peter 2013265920 Dec 11 13:23 /tmp/test.data
-rw-rw   1 peter peter  805306368 Dec 11 13:23 /tmp/test.index

/tmp/test created.
/tmp/test, size=100000000
Load of 100,000,000 records and lookup of 1,000,000 keys took 10.945 seconds

每次查找使用哈希表会更快，因为它是O（1）而不是O（ln），但实现起来更复杂

Answer 2

1. 使用围绕FileInputStream的BufferedInputStream包装的DataInputStream

2. 与每次迭代至少四次系统调用、检查长度和当前大小并执行谁知道有多少次读取来获取字符串和long不同，只需调用readUTF（）和readLong（），直到获得EOFEException