Java8 将一个List转为Map

最新推荐文章于 2025-04-02 07:51:43 发布

李白列夫斯泰

最新推荐文章于 2025-04-02 07:51:43 发布

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分类专栏：数组

原文链接：https://blog.csdn.net/u010979642/article/details/92685846

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将 List 转为 Map<String, T>

public class AnswerApp {
    public static void main(String[] args) throws Exception {
        List<String> names = Lists.newArrayList("Answer", "AnswerAIL", "AI");
        Map<String, Integer> map = names.stream().collect(Collectors.toMap(v -> v, v -> 1));
        System.out.println(map);
    }
}

程序运行输出

{Answer=1, AnswerAIL=1, AI=1}

将 List 转为 Map<K, V>

public static void main(String[] args) throws Exception {
    List<User> users = new ArrayList<>();
    for (int i = 0; i < 3; i++) {
        users.add(new User("answer" + i, new Random().nextInt(100)));
    }
    System.out.println(JSON.toJSONString(users));
	
	System.out.println();
	
    Map<String, Integer> map = users.stream().collect(Collectors.toMap(User::getName, User::getAge));
    System.out.println(map);
}

程序运行输出

[{"age":78,"name":"answer0"},{"age":89,"name":"answer1"},{"age":72,"name":"answer2"}]

{answer2=72, answer1=89, answer0=78}

将 List 转为 Map<String, T>

实现方式1

public class AnswerApp {
    public static void main(String[] args) throws Exception {
        List<User> users = new ArrayList<>();
        for (int i = 0; i < 3; i++) {
        	// 改为此代码, 转map时会报错  Duplicate key User
	//		users.add(new User("answer", new Random().nextInt(100)));
            users.add(new User("answer" + i, new Random().nextInt(100)));
        }
        System.out.println(JSON.toJSONString(users));

        System.out.println();

        Map<String, User> map = users.stream().collect(Collectors.toMap(User::getName, Function.identity()));
        System.out.println(JSON.toJSONString(map));
    }
}

该方式如果 map 的 key(如上述例子的 User::getName 的值) 重复，会抛错java.lang.IllegalStateException: Duplicate key User

实现方式2

public class AnswerApp {

    public static void main(String[] args) throws Exception {
        List<User> users = new ArrayList<>();
        for (int i = 0; i < 3; i++) {
            users.add(new User("answer", new Random().nextInt(100)));
        }
        System.out.println(JSON.toJSONString(users));

        System.out.println();
		
		// 如果 key 重复, 则根据 冲突方法 ·(key1, key2) -> key2· 判断. 解释: key1 key2 冲突时 取 key2 
        Map<String, User> map = users.stream().collect(Collectors.toMap(User::getName, Function.identity(), (key1, key2) -> key2));
        System.out.println(JSON.toJSONString(map));
    }
}

程序运行输出

[{"age":24,"name":"answer"},{"age":89,"name":"answer"},{"age":68,"name":"answer"}]

{"answer":{"age":68,"name":"answer"}}

如果改为 (key1, key2) -> key1 则输出 {"answer":{"age":24,"name":"answer"}}

User 实体

@Data
@NoArgsConstructor
@AllArgsConstructor
public class User {
    private Long id;
    private String name;
    private Integer age;

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

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