Java8 将一个List转为Map

清疚 2023-06-19 11:27 11阅读 0赞

### 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;
        }
    }