集合通过Collectors.toMap转map的注意事项

注意事项

Collectors.toMap
key:不能有两个相同的key,可以为null(多个值对应一个key)
value:可以有相同的value,但value不能为null

不能有两个相同的key

    @Test
    void test0() {
        List<User> userList = new ArrayList<>();
        // 不能有两个相同的key
        userList.add(new User("张三", "18"));
        userList.add(new User("张三", "19"));
        userList.stream().collect(Collectors.toMap(User::getName, User::getAge));
    }

value 不能为null

    @Test
    void test1() {
        List<User> userList = new ArrayList<>();
        // value 不能为null
        userList.add(new User("王五", null));
        userList.stream().collect(Collectors.toMap(User::getName, User::getAge));
    }

key可以为null,value可以重复

    @Test
    void test2() {
        List<User> userList = new ArrayList<>();
        // key可以为null,value可以重复
        userList.add(new User(null, "18"));
        userList.add(new User("李四", "null"));
        userList.add(new User("王五", "null"));
        Map<String, String> nameAgeMap = userList.stream().collect(Collectors.toMap(User::getName, User::getAge));
        System.out.println(nameAgeMap.get(null));
    }

对象不能为null

    @Test
    void test3() {
        List<User> userList = new ArrayList<>();
        // 对象不能为null
        userList.add(null);
        Map<String, String> nameAgeMap = userList.stream().collect(Collectors.toMap(User::getName, User::getAge));
        System.out.println(nameAgeMap.get(null));
    }

解决方式

 userList.stream()
            // null 过滤
            .filter(user -> user != null && user.getAge() != null)
            // 转map,key冲突解决
            .collect(Collectors.toMap(User::getName, User::getAge, (v1, v2) -> v1));

代码

import org.junit.jupiter.api.Test;

import java.util.ArrayList;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;

/**
 * Collectors.toMap
 * key:不能有两个相同的key,可以为null(多个值对应一个key)
 * value:可以有相同的value,但value不能为null
 */
public class MyTest {

    @Test
    void test0() {
        List<User> userList = new ArrayList<>();
        // 不能有两个相同的key
        userList.add(new User("张三", "18"));
        userList.add(new User("张三", "19"));
        userList.stream().collect(Collectors.toMap(User::getName, User::getAge));
    }

    @Test
    void test1() {
        List<User> userList = new ArrayList<>();
        // value 不能为null
        userList.add(new User("王五", null));
        userList.stream().collect(Collectors.toMap(User::getName, User::getAge));
    }

    @Test
    void test2() {
        List<User> userList = new ArrayList<>();
        // key可以为null,value可以重复
        userList.add(new User(null, "18"));
        userList.add(new User("李四", "null"));
        userList.add(new User("王五", "null"));
        Map<String, String> nameAgeMap = userList.stream().collect(Collectors.toMap(User::getName, User::getAge));
        System.out.println(nameAgeMap.get(null));
    }

    @Test
    void test3() {
        List<User> userList = new ArrayList<>();
        // 对象不能为null
        userList.add(null);
        Map<String, String> nameAgeMap = userList.stream().collect(Collectors.toMap(User::getName, User::getAge));
        System.out.println(nameAgeMap.get(null));
    }
    

    class User {

        private String name;

        private String age;

        public User(String name, String age) {
            this.name = name;
            this.age = age;
        }

        public String getName() {
            return name;
        }

        public void setName(String name) {
            this.name = name;
        }

        public String getAge() {
            return age;
        }

        public void setAge(String age) {
            this.age = age;
        }

    }

}

posted @ 2024-08-08 09:29  Toxic-man  阅读(258)  评论(0)    收藏  举报