Java 8 Stream.distinct() 列表去重的操作
作者:Haiyoung
在这篇文章里,我们将提供Java8 Stream distinct()示例。 distinct()返回由该流的不同元素组成的流。distinct()是Stream接口的方法。
distinct()使用hashCode()和equals()方法来获取不同的元素。因此,我们的类必须实现hashCode()和equals()方法。
如果distinct()正在处理有序流,那么对于重复元素,将保留以遭遇顺序首先出现的元素,并且以这种方式选择不同元素是稳定的。
在无序流的情况下,不同元素的选择不一定是稳定的,是可以改变的。distinct()执行有状态的中间操作。
在有序流的并行流的情况下,保持distinct()的稳定性是需要很高的代价的,因为它需要大量的缓冲开销。如果我们不需要保持遭遇顺序的一致性,那么我们应该可以使用通过BaseStream.unordered()方法实现的无序流。
1. Stream.distinct()
distinct()方法的声明如下:
Stream<T> distinct()
它是Stream接口的方法。在此示例中,我们有一个包含重复元素的字符串数据类型列表
DistinctSimpleDemo.java
package com.concretepage;
import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;
public class DistinctSimpleDemo {
public static void main(String[] args) {
List<String> list = Arrays.asList("AA", "BB", "CC", "BB", "CC", "AA", "AA");
long l = list.stream().distinct().count();
System.out.println("No. of distinct elements:"+l);
String output = list.stream().distinct().collect(Collectors.joining(","));
System.out.println(output);
}
}
Output
No. of distinct elements:3
AA,BB,CC
2. Stream.distinct() with List of Objects
在此示例中,我们有一个Book对象列表。 为了对列表进行去重,该类将重写hashCode()和equals()。
Book.java
package com.concretepage;
public class Book {
private String name;
private int price;
public Book(String name, int price) {
this.name = name;
this.price = price;
}
public String getName() {
return name;
}
public int getPrice() {
return price;
}
@Override
public boolean equals(final Object obj) {
if (obj == null) {
return false;
}
final Book book = (Book) obj;
if (this == book) {
return true;
} else {
return (this.name.equals(book.name) && this.price == book.price);
}
}
@Override
public int hashCode() {
int hashno = 7;
hashno = 13 * hashno + (name == null ? 0 : name.hashCode());
return hashno;
}
}
DistinctWithUserObjects.java
package com.concretepage;
import java.util.ArrayList;
import java.util.List;
public class DistinctWithUserObjects {
public static void main(String[] args) {
List<Book> list = new ArrayList<>();
{
list.add(new Book("Core Java", 200));
list.add(new Book("Core Java", 200));
list.add(new Book("Learning Freemarker", 150));
list.add(new Book("Spring MVC", 300));
list.add(new Book("Spring MVC", 300));
}
long l = list.stream().distinct().count();
System.out.println("No. of distinct books:"+l);
list.stream().distinct().forEach(b -> System.out.println(b.getName()+ "," + b.getPrice()));
}
}
Output
No. of distinct books:3 Core Java,200 Learning Freemarker,150 Spring MVC,300
3. Distinct by Property
distinct()不提供按照属性对对象列表进行去重的直接实现。它是基于hashCode()和equals()工作的。
如果我们想要按照对象的属性,对对象列表进行去重,我们可以通过其它方法来实现。
如下代码段所示:
static <T> Predicate<T> distinctByKey(Function<? super T, ?> keyExtractor) {
Map<Object,Boolean> seen = new ConcurrentHashMap<>();
return t -> seen.putIfAbsent(keyExtractor.apply(t), Boolean.TRUE) == null;
}
上面的方法可以被Stream接口的 filter()接收为参数,如下所示:
list.stream().filter(distinctByKey(b -> b.getName()));
distinctByKey()方法返回一个使用ConcurrentHashMap 来维护先前所见状态的 Predicate 实例,如下是一个完整的使用对象属性来进行去重的示例。
DistinctByProperty.java
package com.concretepage;
import java.util.ArrayList;
import java.util.List;
import java.util.Map;
import java.util.concurrent.ConcurrentHashMap;
import java.util.function.Function;
import java.util.function.Predicate;
public class DistinctByProperty {
public static void main(String[] args) {
List<Book> list = new ArrayList<>();
{
list.add(new Book("Core Java", 200));
list.add(new Book("Core Java", 300));
list.add(new Book("Learning Freemarker", 150));
list.add(new Book("Spring MVC", 200));
list.add(new Book("Hibernate", 300));
}
list.stream().filter(distinctByKey(b -> b.getName()))
.forEach(b -> System.out.println(b.getName()+ "," + b.getPrice()));
}
private static <T> Predicate<T> distinctByKey(Function<? super T, ?> keyExtractor) {
Map<Object,Boolean> seen = new ConcurrentHashMap<>();
return t -> seen.putIfAbsent(keyExtractor.apply(t), Boolean.TRUE) == null;
}
}
Output
Core Java,200 Learning Freemarker,150 Spring MVC,200 Hibernate,300
from : https://www.concretepage.com/java/jdk-8/java-8-distinct-example
补充知识:List集合常规去重与java8新特性去重方法
一、常规去重
碰到List去重的问题,除了遍历去重,我们常常想到利用Set集合不允许重复元素的特点,通过List和Set互转,来去掉重复元素。
// 遍历后判断赋给另一个list集合,保持原来顺序
public static void ridRepeat1(List<String> list) {
System.out.println("list = [" + list + "]");
List<String> listNew = new ArrayList<String>();
for (String str : list) {
if (!listNew.contains(str)) {
listNew.add(str);
}
}
System.out.println("listNew = [" + listNew + "]");
}
// set集合去重,保持原来顺序
public static void ridRepeat2(List<String> list) {
System.out.println("list = [" + list + "]");
List<String> listNew = new ArrayList<String>();
Set set = new HashSet();
for (String str : list) {
if (set.add(str)) {
listNew.add(str);
}
}
System.out.println("listNew = [" + listNew + "]");
}
// Set去重 由于Set的无序性,不会保持原来顺序
public static void ridRepeat3(List<String> list) {
System.out.println("list = [" + list + "]");
Set set = new HashSet();
List<String> listNew = new ArrayList<String>();
set.addAll(list);
listNew.addAll(set);
System.out.println("listNew = [" + listNew + "]");
}
// Set去重(将ridRepeat3方法缩减为一行) 无序
public static void ridRepeat4(List<String> list) {
System.out.println("list = [" + list + "]");
List<String> listNew = new ArrayList<String>(new HashSet(list));
System.out.println("listNew = [" + listNew + "]");
}
// Set去重并保持原先顺序
public static void ridRepeat5(List<String> list) {
System.out.println("list = [" + list + "]");
List<String> listNew2= new ArrayList<String>(new LinkedHashSet<String>(list));
System.out.println("listNew = [" + listNew + "]");
}
二、java8的stream写法实现去重
1、distinct去重
//利用java8的stream去重 List uniqueList = list.stream().distinct().collect(Collectors.toList()); System.out.println(uniqueList.toString());
distinct()方法默认是按照父类Object的equals与hashCode工作的。所以:
上面的方法在List元素为基本数据类型及String类型时是可以的,但是如果List集合元素为对象,却不会奏效。不过如果你的实体类对象使用了目前广泛使用的lombok插件相关注解如:@Data,那么就会自动帮你重写了equals与hashcode方法,当然如果你的需求是根据某几个核心字段属性判断去重,那么你就要在该类中自定义重写equals与hashcode方法了。
2、也可以通过新特性简写方式实现
不过该方式不能保持原列表顺序而是使用了TreeSet按照字典顺序排序后的列表,如果需求不需要按原顺序则可直接使用。
//根据name属性去重
List<User> lt = list.stream().collect(
collectingAndThen(
toCollection(() -> new TreeSet<>(Comparator.comparing(User::getName))), ArrayList::new));
System.out.println("去重后的:" + lt);
//根据name与address属性去重
List<User> lt1 = list.stream().collect(
collectingAndThen(
toCollection(() -> new TreeSet<>(Comparator.comparing(o -> o.getName() + ";" + o.getAddress()))), ArrayList::new));
System.out.println("去重后的:" + lt);
当需求中明确有排序要求也可以按上面简写方式再次加工处理使用stream流的sorted()相关API写法。
List<User> lt = list.stream().collect(
collectingAndThen(
toCollection(() -> new TreeSet<>(Comparator.comparing(User::getName))),v -> v.stream().sorted().collect(Collectors.toList())));
3、通过 filter() 方法
我们首先创建一个方法作为 Stream.filter() 的参数,其返回类型为 Predicate,原理就是判断一个元素能否加入到 Set 中去,代码如下:
private static <T> Predicate<T> distinctByKey(Function<? super T, ?> keyExtractor) {
Set<Object> seen = ConcurrentHashMap.newKeySet();
return t -> seen.add(keyExtractor.apply(t));
}
使用如下:
@Test
public void distinctByProperty() throws JsonProcessingException {
// 这里第二种方法我们通过过滤来实现根据对象某个属性去重
ObjectMapper objectMapper = new ObjectMapper();
List<Student> studentList = getStudentList();
System.out.print("去重前 :");
System.out.println(objectMapper.writeValueAsString(studentList));
studentList = studentList.stream().distinct().collect(Collectors.toList());
System.out.print("distinct去重后:");
System.out.println(objectMapper.writeValueAsString(studentList));
// 这里我们将 distinctByKey() 方法作为 filter() 的参数,过滤掉那些不能加入到 set 的元素
studentList = studentList.stream().filter(distinctByKey(Student::getName)).collect(Collectors.toList());
System.out.print("根据名字去重后 :");
System.out.println(objectMapper.writeValueAsString(studentList));
}
去重前:
[{"stuNo":"001","name":"Tom"},{"stuNo":"001","name":"Tom"},{"stuNo":"003","name":"Tom"}]
distinct去重后:
[{"stuNo":"001","name":"Tom"},{"stuNo":"003","name":"Tom"}]
根据名字去重后 :
[{"stuNo":"001","name":"Tom"}]
三、相同元素累计求和等操作
除了集合去重意外,工作中还有一种常见的需求,例如:在所有商品订单中,计算同一家店铺不同商品名称的商品成交额,可以直接通过sql语句获取,这里写一下如何通过java简单实现。举一个类似的案例:计算相同姓名与住址的用户年龄之和。
User.java
package com.example.demo.dto;
import java.io.Serializable;
import java.util.Objects;
/**
* @author: shf
* description:
* date: 2019/10/30 10:21
*/
public class User implements Serializable {
private static final long serialVersionUID = 1L;
private Long id;
private String name;
private String address;
private Integer age;
public User() {
}
public User(String name, String address, Integer age) {
this.name = name;
this.address = address;
this.age = age;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getAddress() {
return address;
}
public void setAddress(String address) {
this.address = address;
}
public Integer getAge() {
return age;
}
public void setAge(Integer age) {
this.age = age;
}
@Override
public String toString() {
return "User{" +
"name='" + name + '\'' +
", address='" + address + '\'' +
", age=" + age +
'}';
}
@Override
public boolean equals(Object obj) {
if (this == obj) {
return true;//地址相等
}
if (obj == null) {
return false;//非空性:对于任意非空引用x,x.equals(null)应该返回false。
}
if (obj instanceof User) {
User other = (User) obj;
//需要比较的字段相等,则这两个对象相等
if (Objects.equals(this.name, other.name)
&& Objects.equals(this.address, other.address)) {
return true;
}
}
return false;
}
@Override
public int hashCode() {
return Objects
.hash(name, address);
}
}
测试代码:
package com.example.demo;
import com.example.demo.dto.User;
import java.util.*;
import java.util.stream.Collectors;
public class FirCes {
public static void main(String[] args) {
/*构建测试数据集合*/
User user1 = new User("a小张1", "a1", 10);
User user2 = new User("b小张2", "a2", 10);
User user3 = new User("c小张3", "a3", 10);
User user3_3 = new User("c小张3", "a", 10);
User user33 = new User("c小张3", "a3", 10);
User user4 = new User("d小张4", "a4", 10);
User user5 = new User("e小张5", "a5", 10);
List<User> list = new ArrayList<>();
list.add(user1);
list.add(user2);
list.add(user3);
list.add(user3_3);
list.add(user33);
list.add(user4);
list.add(user5);
//按相同name与address属性分组User用户
Map<User, List<User>> listMap = list.stream().collect(Collectors.groupingBy(v -> v));
/*先看一下分组效果*/
listMap.forEach((key, value) -> {
System.out.println("========");
System.out.println("key:" + key);
value.forEach(obj -> {
System.out.println(obj);
});
});
/*最终执行结果*/
List<User> listNew = listMap.keySet().stream().map(u -> {
int sum = listMap.get(u).stream().mapToInt(i -> i.getAge()).sum();
//需要注意的是:这里也会改变原list集合中的原数据。因为这里的u分组时就是来自原集合中的一个地址对象,
// 即:指向了原集合中的一个对象的地址。如果不想原集合被影响,这里可以new User()新的对象赋值并返回新对象
u.setAge(sum);
return u;
}).collect(Collectors.toList());
System.out.println("listNew:" + listNew);
System.err.println("list:" + list);
//但是一个实体类只能重写一次equals方法,如果有多种判别需求就不好满足了,
// 可以定义多个不同类名相同属性的类或者下面这种方式解决
Map<String, List<User>> listMap1 = list.stream().collect(Collectors
.groupingBy(v -> Optional.ofNullable(v.getName()).orElse("") + "_" + Optional.ofNullable(v.getAddress()).orElse("")));
/*先看一下分组效果*/
listMap1.forEach((key, value) -> {
System.out.println("========");
System.out.println("key:" + key);
value.forEach(obj -> {
System.out.println(obj);
});
});
/*最终执行结果*/
List<User> listNew1 = listMap1.keySet().stream().map(u -> {
int sum = listMap1.get(u).stream().mapToInt(i -> i.getAge()).sum();
User user = listMap1.get(u).get(0);
//这里和上面一样的原理,也会影响原list集合中的被指向的地址的对象数据
user.setAge(sum);
return user;
}).collect(Collectors.toList());
System.out.println("listNew1:" + listNew1);
System.err.println("list:" + list);
}
}
打印日志:
========
key:User{name='b小张2', address='a2', age=10}
User{name='b小张2', address='a2', age=10}
========
key:User{name='c小张3', address='a', age=10}
User{name='c小张3', address='a', age=10}
========
key:User{name='c小张3', address='a3', age=10}
User{name='c小张3', address='a3', age=10}
User{name='c小张3', address='a3', age=10}
========
key:User{name='a小张1', address='a1', age=10}
User{name='a小张1', address='a1', age=10}
========
key:User{name='d小张4', address='a4', age=10}
User{name='d小张4', address='a4', age=10}
========
key:User{name='e小张5', address='a5', age=10}
User{name='e小张5', address='a5', age=10}
listNew:[User{name='b小张2', address='a2', age=10}, User{name='c小张3', address='a', age=10}, User{name='c小张3', address='a3', age=20}, User{name='a小张1', address='a1', age=10}, User{name='d小张4', address='a4', age=10}, User{name='e小张5', address='a5', age=10}]
list:[User{name='a小张1', address='a1', age=10}, User{name='b小张2', address='a2', age=10}, User{name='c小张3', address='a3', age=20}, User{name='c小张3', address='a', age=10}, User{name='c小张3', address='a3', age=10}, User{name='d小张4', address='a4', age=10}, User{name='e小张5', address='a5', age=10}]
========
key:a小张1_a1
User{name='a小张1', address='a1', age=10}
========
key:c小张3_a
User{name='c小张3', address='a', age=10}
========
key:d小张4_a4
User{name='d小张4', address='a4', age=10}
========
key:e小张5_a5
User{name='e小张5', address='a5', age=10}
========
key:b小张2_a2
User{name='b小张2', address='a2', age=10}
========
key:c小张3_a3
User{name='c小张3', address='a3', age=20}
User{name='c小张3', address='a3', age=10}
listNew1:[User{name='a小张1', address='a1', age=10}, User{name='c小张3', address='a', age=10}, User{name='d小张4', address='a4', age=10}, User{name='e小张5', address='a5', age=10}, User{name='b小张2', address='a2', age=10}, User{name='c小张3', address='a3', age=30}]
list:[User{name='a小张1', address='a1', age=10}, User{name='b小张2', address='a2', age=10}, User{name='c小张3', address='a3', age=30}, User{name='c小张3', address='a', age=10}, User{name='c小张3', address='a3', age=10}, User{name='d小张4', address='a4', age=10}, User{name='e小张5', address='a5', age=10}]
Process finished with exit code 0
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