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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<String, Integer> map;//索引类型和bitSet在bsList中下标的映射关系 private List<String> list;//索引类型的值集合,例如gender:girl,boy private List<BitSet> bsList;//bitset集合 public BitSetIndex() { } public BitSetIndexModel(String type) { this.type = type; map = new ConcurrentHashMap<String, Integer>(); list = new ArrayList<String>(); bsList = new ArrayList<BitSet>(); } public String getType() { return type; } public void setType(String type) { this.type = type; } public Map<String, Integer> getMap() { return map; } public void setMap(ConcurrentHashMap<String, Integer> map) { this.map = map; } public List<String> getList() { return list; } public void setList(List<String> list) { this.list = list; } public List<BitSet> getBsList() { return bsList; } public void setBsList(List<BitSet> 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<String> 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<Integer> getRealIndexs(BitSet bitset) { List<Integer> indexs = new ArrayList<Integer>(); 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<Integer, User> 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<Integer, User>(); } /** * 构建索引 * @param users */ public void createIndex(List<User> 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<User> 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<Integer> 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<User> buildData() { String[] addresss = { "北京", "上海", "深圳" }; String[] ages = { "16", "17", "18" }; List<User> 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
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