这篇文章主要为大家展示了“如何使用bitset实现毫秒级查询”，内容简而易懂，条理清晰，希望能够帮助大家解决疑惑，下面让小编带领大家一起研究并学习一下“如何使用bitset实现毫秒级查询”这篇文章吧。
成都创新互联专注于企业成都全网营销推广、网站重做改版、会昌网站定制设计、自适应品牌网站建设、HTML5建站、成都商城网站开发、集团公司官网建设、成都外贸网站建设、高端网站制作、响应式网页设计等建站业务，价格优惠性价比高，为会昌等各大城市提供网站开发制作服务。
bitset介绍
看JDK中的解释简直一头雾水，用我自己的理解概括一下
1.bitset的内部实现是long数组
2.set中每一个位的默认值为false（0）
3.bitset长度按需增长
4.bitset非线程安全
bitset关键方法分析
/**
* Sets the bit at the specified index to {@code true}.
*
* @param bitIndex a bit index
* @throws IndexOutOfBoundsException if the specified index is negative
* @since JDK1.0
*/
public void set(int bitIndex) {
if (bitIndex < 0)
throw new IndexOutOfBoundsException("bitIndex < 0: " + bitIndex);
int wordIndex = wordIndex(bitIndex);
expandTo(wordIndex);
words[wordIndex] |= (1L << bitIndex); // Restores invariants
checkInvariants();
}设置指定“位”为true，bitIndex参数为非负整数。假设我们执行以下代码,观察上面代码中worIndex，words[wordIndex]值的变化
BitSet bs = new BitSet() bs.set(0); bs.set(1); bs.set(2); bs.set(3); bs.set(4);
| bitIndex | wordIndex | words[wordIndex] | words[wordIndex]二进制表示 |
|---|---|---|---|
| 0 | 0 | 1 | 0001 |
| 1 | 0 | 3 | 0011 |
| 2 | 0 | 7 | 0111 |
| 3 | 0 | 15 | 1111 |
| 4 | 0 | 31 | 0001 1111 |
通过上表，我们可以很清晰的根据bitIndex和words[wordIndex]二进制值的对应关系，得到一个结论，即：bitset中每一个long可以表示64个非负整数在bitSet中存在与否。例如：0001可以表示整数0在bitset中存在，1111可以表示整数3,2,1,0在bitset中存在。
进入正题，实现bitset毫秒级查询
想象一个场景，我们有一张user表
CREATE TABLE `user` ( `id` int(11) NOT NULL AUTO_INCREMENT, `name` varchar(255) NOT NULL, `address` varchar(255) NOT NULL comment '地址', `gender` varchar(10) NOT NULL comment '性别', `age` varchar(10) NOT NULL, PRIMARY KEY (`uid`) ) ENGINE=InnoDB AUTO_INCREMENT=0 DEFAULT CHARSET=utf8;
假设我们要查询“北京市18岁的女生”，那么对应的sql如下：
select * from `user` where address='北京' and age='18' and gender='girl';
如何使用bitset实现同样的查询呢？
1.将user表数据加载进内存中
2.为user表建立address,age,gender维度的bitset索引
3.根据索引查询数据
1.将user表数据加载进内存中
将user表从数据库读取出来存入List
User实体类：
public class User implements Cloneable {
private String name;
private String address;
private String gender;
private String age;
@Override
public String toString() {
return "User [name=" + name + ", address=" + address + ", gender=" + gender + ", age=" + age + "]";
}
@Override
public User clone() {
User user = null;
try {
user = (User) super.clone();
} catch (CloneNotSupportedException e) {
e.printStackTrace();
}
return user;
}
//省略get set 方法。。。2.建立索引
创建bitset索引模型类
public class BitSetIndexModel {
private String type;//索引类型：address，age，gender
private ConcurrentHashMap map;//索引类型和bitSet在bsList中下标的映射关系
private List list;//索引类型的值集合，例如gender：girl，boy
private List bsList;//bitset集合
public BitSetIndex() {
}
public BitSetIndexModel(String type) {
this.type = type;
map = new ConcurrentHashMap();
list = new ArrayList();
bsList = new ArrayList();
}
public String getType() {
return type;
}
public void setType(String type) {
this.type = type;
}
public Map getMap() {
return map;
}
public void setMap(ConcurrentHashMap map) {
this.map = map;
}
public List getList() {
return list;
}
public void setList(List list) {
this.list = list;
}
public List getBsList() {
return bsList;
}
public void setBsList(List bsList) {
this.bsList = bsList;
}
/**
*
* @param str
* @param i
*/
public void createIndex(String str, int i) {
BitSet bitset = null;
//获取‘str'对应的bitset在bsList中的下标
Integer index = this.getMap().get(str);
if (index != null) {
bitset = this.getBsList().get(index);
if (bitset == null) {
bitset = new BitSet();
this.getBsList().add(index, bitset);
}
bitset.set(i, true);// 将str对应的位置为true,true可省略
} else {
bitset = new BitSet();
List list = this.getList();
list.add(str);
index = list.size() - 1;
bitset.set(i, true);
this.getBsList().add(bitset);
this.getMap().put(str, index);
}
}
/**
* 从entity里拿出符合条件的bitset
*
* @param str
* @return
*/
public BitSet get(String str) {
BitSet bitset = null;
str = str.toLowerCase();
Integer index = this.getMap().get(str);
if (index != null) {
bitset = this.getBsList().get(index);
} else {
bitset = new BitSet();
}
return bitset;
}
/**
* bitset的与运算
*
* @param str
* @param bitset
* @return
*/
public BitSet and(String str, BitSet bitset) {
if (str != null) {
str = str.toLowerCase();
if (bitset != null) {
bitset.and(get(str));
} else {
bitset = new BitSet();
bitset.or(get(str));
}
}
return bitset;
}
/**
* bitset的或运算
*
* @param str
* @param bitset
* @return
*/
public BitSet or(String str, BitSet bitset) {
if (str != null) {
str = str.toLowerCase();
if (bitset != null) {
bitset.or(get(str));
} else {
bitset = new BitSet();
bitset.or(get(str));
}
}
return bitset;
}
/**
* 获取bitset值为true的 即 把 bitset翻译为list的索引
*
* @param bitset
* @return
*/
public static List getRealIndexs(BitSet bitset) {
List indexs = new ArrayList();
if (bitset != null) {
int i = bitset.nextSetBit(0);
if (i != -1) {
indexs.add(i);
for (i = bitset.nextSetBit(i + 1); i >= 0; i = bitset.nextSetBit(i + 1)) {
int endOfRun = bitset.nextClearBit(i);
do {
indexs.add(i);
} while (++i < endOfRun);
}
}
}
return indexs;
}
} 为每一个user对象创建address,gender,age维度索引
public class UserIndexStore {
private static final String ADDRESS = "address";
private static final String GENDER = "gender";
private static final String AGE = "age";
private BitSetIndexModel address;
private BitSetIndexModel gender;
private BitSetIndexModel age;
private ConcurrentHashMap userMap;//存储所有的user数据
public static final UserIndexStore INSTANCE = getInstance();
private UserIndexStore() {
init();
}
public static UserIndexStore getInstance() {
return UserIndexStoreHolder.instance;
}
private static class UserIndexStoreHolder {
private static UserIndexStore instance = new UserIndexStore();
}
private void init() {
this.address = new BitSetIndexModel(ADDRESS);
this.gender = new BitSetIndexModel(GENDER);
this.age = new BitSetIndexModel(AGE);
userMap = new ConcurrentHashMap();
}
/**
* 构建索引
* @param users
*/
public void createIndex(List users) {
if (users != null && users.size() > 0) {
for (int index = 0; index < users.size(); index++) {
User user = users.get(index);
getAddress().update(user.getAddress(), index);
getGender().update(user.getGender(), index);
getAge().update(user.getAge(), index);
this.userMap.put(index, user);
}
}
}
public BitSet query(String address, String gender, String age) {
BitSet bitset = null;
bitset = getAddress().and(address, bitset);
bitset = getGender().and(gender, bitset);
bitset = getAge().and(age, bitset);
return bitset;
}
public User findUser(Integer index) {
User user = this.userMap.get(index);
if (user != null) {
return user.clone();//可能会对user做修改操作,要保证内存原始数据不变
}
return null;
}
public BitSetIndexModel getAddress() {
return address;
}
public void setAddress(BitSetIndexModel address) {
this.address = address;
}
public BitSetIndexModel getGender() {
return gender;
}
public void setGender(BitSetIndexModel gender) {
this.gender = gender;
}
public BitSetIndexModel getAge() {
return age;
}
public void setAge(BitSetIndexModel age) {
this.age = age;
}
} 3.测试bitset
public class BitSetTest {
public static void main(String[] args) {
List users = buildData();
UserIndexStore.getInstance().createIndex(users);
ExecutorService executorService = Executors.newFixedThreadPool(20);
int num = 2000;
long begin1 = System.currentTimeMillis();
for (int i = 0; i < num; i++) {
Runnable syncRunnable = new Runnable() {
@Override
public void run() {
List indexs = BitSetIndexModel.getRealIndexs(UserIndexStore.getInstance().query("北京", "girl", "18"));
for (Integer index : indexs) {
UserIndexStore.getInstance().findUser(index);
}
}
};
executorService.execute(syncRunnable);
}
executorService.shutdown();
while (true) {
try {
if (executorService.awaitTermination(1, TimeUnit.SECONDS)) {
System.out.println("单次查询时间为：" + (System.currentTimeMillis() - begin1) / num + "ms");
break;
}
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
private static List buildData() {
String[] addresss = { "北京", "上海", "深圳" };
String[] ages = { "16", "17", "18" };
List users = new ArrayList<>();
for (int i = 0; i < 200000; i++) {
User user = new User();
user.setName("name" + i);
int rand = ThreadLocalRandom.current().nextInt(3);
user.setAddress(addresss[ThreadLocalRandom.current().nextInt(3)]);
user.setGender((rand & 1) == 0 ? "girl" : "boy");
user.setAge(ages[ThreadLocalRandom.current().nextInt(3)]);
users.add(user);
}
return users;
}
} 测试结果(查询2w次)：
数据量（users.size()） | 并发数 | 平均查询时间
---|---|---
20w | 10 | 1ms
50w | 20 | 3ms
100w| 50 | 9ms
测试机为thinkpad x240 i5 8g内存
4.总结
==优点==：
通过测试发现随着数据量的增大和并发数的提高，平均耗时并没有明显升高，并且响应时间都在10ms以内
==缺点==：
1.不适合数据量太大的情况,因为需要把数据全部加载进内存
2.不适合复杂查询
3.不适合对name,id等唯一值做查询
以上是“如何使用bitset实现毫秒级查询”这篇文章的所有内容，感谢各位的阅读！相信大家都有了一定的了解，希望分享的内容对大家有所帮助，如果还想学习更多知识，欢迎关注创新互联行业资讯频道！