我有一个1.7G文件,格式如下:
String Long String Long String Long String Long ... etc
基本上, String 是一个键, Long 是一个hashmap中的值,我有兴趣在我的应用程序中运行其他任何东西之前进行初始化。
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分钟才能完成,我确信有更好的方法可以做到这一点,所以我会感激任何帮助或指针。
感谢
按EJP建议更新?
EJP感谢您的建议,我希望这就是您的意思。如果这是错误的话,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();
}
}
答案 0 :(得分:4)
使用围绕FileInputStream的BufferedInputStream包装的DataInputStream。
而不是每次迭代至少四次系统调用,检查长度和当前大小,并执行谁知道获取字符串和long的读取次数,只需调用readUTF()和readLong()直到你得到EOFException。
答案 1 :(得分:2)
我会构造文件,以便它可以在适当的位置使用。即不加载这种方式。由于您有可变长度记录,您可以构造每个记录的位置数组,然后按顺序放置密钥,以便您可以对数据执行二进制搜索。 (或者您可以使用自定义哈希表)然后可以使用方法将其包装起来,该方法隐藏数据实际存储在文件中而不是变成数据对象的事实。
如果你做了这一切,“加载”阶段变得多余,你不需要创建这么多的对象。
这是一个很长的例子,但希望能说明可能的事情。
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2 GB的原始数据并执行一百万次查找。它以这样的方式编写,即加载和查找使用非常少的堆。 (&lt;&lt;&lt; 1 MB)
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
使用哈希表查找每次查找会更快,因为它是O(1)而不是O(ln N),但实现起来更复杂。