代码收藏家 Java实现快速去重：基于对象某一属性的去重方法指南
一、导言
🧐📚 Java中的对象去重操作？跟着小编一起学习吧！👇
在处理对象集合时，有时候我们需要根据对象的某个属性进行去重操作。Java给我们提供了多种方法来实现这个功能。今天，小编就来给大家介绍一下如何使用Java根据对象的某个属性进行去重操作。💫
方案一：使用自定义equals()和hashCode()方法
I.原理讲解
II. 代码示例
// 重新equals和hashCode方法
public class Person {
public String name;
public Integer age;
public Sex sex;
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Person person = (Person) o;
return Objects.equals(name, person.name);
}
@Override
public int hashCode() {
return Objects.hash(name);
}
······
}
public class ObjectDistinctTest {
public Person[] peoples;
public List<Person> personList;
// 构建数据源
@Before
public void before() {
Person person = new Chinese("大白", 21, Sex.man);
Person person1 = new Chinese("小兰", 25, Sex.woman);
Person person2 = new Chinese("小红", 23, Sex.woman);
Person person3 = new Chinese("小赵", 20, Sex.man);
Person person4 = new Chinese("小青", 18, Sex.woman);
Person person5 = new Chinese("小赵", 17, Sex.man);
Person person6 = new Chinese("小麻", 50, Sex.man);
this.peoples =new Person[]{ person, person1, person2, person3, person4, person5, person6};
this.personList = Arrays.asList(peoples);
}
/**
* 使用自定义equals()和hashCode()方法
*/
@Test
public void testCase01() {
// 需求：根据Person对象的name属性进行去重,name属性相同，则视为重复元素
// 存入顺序和取出顺序不一致
Set<Person> distinctPersonSet = new HashSet<>();
for (Person person : personList) {
distinctPersonSet.add(person); //
}
List<Person> distinctPersonList = new ArrayList<>(distinctPersonSet);
distinctPersonList.forEach(System.out::println); // 打印去重后的数据
}
/**
* 使用自定义equals()和hashCode()方法
*/
@Test
public void testCase02() {
// 或使用 LinkedHashSet 保持插入顺序
Set<Person> distinctPersonSet = new LinkedHashSet<>();
for (Person person : personList) {
distinctPersonSet.add(person);
}
List<Person> distinctPersonList = new ArrayList<>(distinctPersonSet);
distinctPersonList.forEach(System.out::println); // 打印去重后的数据
}
}
方案二：使用循环和集合
I.原理讲解
II. 代码示例
public class ObjectDistinctTest {
public Person[] peoples;
public List<Person> personList;
// 构建数据源
@Before
public void before() {
Person person = new Chinese("大白", 21, Sex.man);
Person person1 = new Chinese("小兰", 25, Sex.woman);
Person person2 = new Chinese("小红", 23, Sex.woman);
Person person3 = new Chinese("小赵", 20, Sex.man);
Person person4 = new Chinese("小青", 18, Sex.woman);
Person person5 = new Chinese("小赵", 17, Sex.man);
Person person6 = new Chinese("小麻", 50, Sex.man);
this.peoples =new Person[]{ person, person1, person2, person3, person4, person5, person6};
this.personList = Arrays.asList(peoples);
}
@Test
public void testCase() {
// 此方式无需实现equals方法
List<Person> newList = new ArrayList<>();
for (Person person: this.personList
) {
if (newList.stream().noneMatch(x -> x.getName().equals(person.getName()))) { // 判断新集合中是否包含此对象，不包含才能加入到新集合中
newList.add(person);
}
}
List<Person> distinctList = newList.stream().sorted(Comparator.comparing(Person::getAge)).collect(Collectors.toList());
distinctList.forEach(System.out::println);
}
}
方案三：使用自定义equals()和Java 8中的distinct方法
I.原理讲解
II. 代码示例
// 重新equals和hashCode方法
public class Person {
public String name;
public Integer age;
public Sex sex;
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Person person = (Person) o;
return Objects.equals(name, person.name);
}
@Override
public int hashCode() {
return Objects.hash(name);
}
······
}
public class ObjectDistinctTest {
public Person[] peoples;
public List<Person> personList;
// 构建数据源
@Before
public void before() {
Person person = new Chinese("大白", 21, Sex.man);
Person person1 = new Chinese("小兰", 25, Sex.woman);
Person person2 = new Chinese("小红", 23, Sex.woman);
Person person3 = new Chinese("小赵", 20, Sex.man);
Person person4 = new Chinese("小青", 18, Sex.woman);
Person person5 = new Chinese("小赵", 17, Sex.man);
Person person6 = new Chinese("小麻", 50, Sex.man);
this.peoples =new Person[]{ person, person1, person2, person3, person4, person5, person6};
this.personList = Arrays.asList(peoples);
}
/**
* 使用自定义equals()和Stream API中的distinct方法实现元素的去重
*/
@Test
public void testCase01() {
// 需求：根据Person对象的name属性进行去重,name属性相同，则视为重复元素
this.personList.stream() // 获取流对象
.distinct() // 去除重复元素
.collect(Collectors.toList()) // 收集为List集合
.forEach(System.out::println); // 打印去重后的元素
}
}
方案四：将List以name属性为key转换为Map
I.原理讲解
II. 代码示例
public class ObjectDistinctTest {
public Person[] peoples;
public List<Person> personList;
// 构建数据源
@Before
public void before() {
Person person = new Chinese("大白", 21, Sex.man);
Person person1 = new Chinese("小兰", 25, Sex.woman);
Person person2 = new Chinese("小红", 23, Sex.woman);
Person person3 = new Chinese("小赵", 20, Sex.man);
Person person4 = new Chinese("小青", 18, Sex.woman);
Person person5 = new Chinese("小赵", 17, Sex.man);
Person person6 = new Chinese("小麻", 50, Sex.man);
this.peoples =new Person[]{ person, person1, person2, person3, person4, person5, person6};
this.personList = Arrays.asList(peoples);
}
@Test
public void testCase03() {
this.personList.stream() // 获取流对象
.collect(Collectors.toMap(Person::getName, y -> y,(s, a) -> s))// 将List根据Person类中的name属性收集为Map，key为name，value为Person对象，当后面对象name属性跟前者相同时，使用前者。
.values() // 获取Map中的key
.stream()
.collect(Collectors.toList())
.forEach(System.out::println);
}
}五、总结
想象一下，现在你可以轻松地根据对象的某一属性去除重复对象了！🚀
不仅如此，这个技巧还可以应用在各种不同的场景中，让你的代码更加简洁高效！💪
不管是初学者还是资深开发者，都会对这个技巧爱不释手！快来试试吧！😍
