Java对象集合按某个字段转为map

╰+哭是因爲堅強的太久メ 2023-10-02 17:47 33阅读 0赞

## 前言 ##

在开发中，很多场景下，聪数据源获取到了对象的集合数据后，需要对这部分的数据进行转化，其中按照某个字段转化为 <字段,对象> 或者 <字段,对象集合>的需求非常多，下面来看一个具体的案例吧

## 传统实现 ##

对象类

class User {
        private String userId;
        private String userName;
        private int age;
    
        @Override
        public String toString() {
            return "User{" +
                    "userId='" + userId + '\'' +
                    ", userName='" + userName + '\'' +
                    ", age=" + age +
                    '}';
        }
    
        public User() {
        }
    
        public User(String userId, String userName, int age) {
            this.userId = userId;
            this.userName = userName;
            this.age = age;
        }
    
        public String getUserId() {
            return userId;
        }
    
        public void setUserId(String userId) {
            this.userId = userId;
        }
    
        public String getUserName() {
            return userName;
        }
    
        public void setUserName(String userName) {
            this.userName = userName;
        }
    
        public int getAge() {
            return age;
        }
    
        public void setAge(int age) {
            this.age = age;
        }
    }

**1、对象列表转为 <key，对象>**

传统实现

public class Utils {
    
        public static void main(String[] args) {
    
            User user1 = new User("001","小张",25);
            User user2 = new User("002","小王",15);
            User user3 = new User("003","小黄",19);
    
            List<User> userList = new ArrayList<>();
            userList.add(user1);
            userList.add(user2);
            userList.add(user3);
    
            Map<String,User> userMap = new HashMap<>();
            for(User user : userList){
                userMap.put(user.getUserId(),user);
            }
            System.out.println(userMap);
    
    
        }

![watermark_type_d3F5LXplbmhlaQ_shadow_50_text_Q1NETiBA6YCG6aOO6aOe57-U55qE5bCP5Y-U_size_20_color_FFFFFF_t_70_g_se_x_16][]

通过打印输出的控制台结果，可以实现 <key,对象>结果的获取，这是使用传统的方式实现，对大多数同学来说，这种实现也是一种常规比较容易的方式

**Java8实现**

public static void main(String[] args) {
    
            User user1 = new User("001","小张",25);
            User user2 = new User("002","小王",15);
            User user3 = new User("003","小黄",19);
    
            List<User> userList = new ArrayList<>();
            userList.add(user1);
            userList.add(user2);
            userList.add(user3);
    
    
            Map<String, User> stringUserMap = toMap(userList, User::getUserId);
            System.out.println(stringUserMap);
    
        }
    
        /**
         * list转Map 数据格式为<String,E>
         * key为指定的字段，value 表示当前字段
         */
        public static <K, E> Map<K, E> toMap(Collection<E> list, Function<E, K> function) {
            return list.stream().collect(Collectors.toMap(function, a -> a, (k1, K2) -> k1));
        }

Java8提供了很多函数式接口可供开发中调用，比如上面的这个需求就可以直接调用相关的函数进行操作，可以得到同样的输出结果

![watermark_type_d3F5LXplbmhlaQ_shadow_50_text_Q1NETiBA6YCG6aOO6aOe57-U55qE5bCP5Y-U_size_20_color_FFFFFF_t_70_g_se_x_16 1][]

上面的实现方式，适合那些对象中某个字段是唯一的场景

**2、对象列表按照字段进行分组**

public static void main(String[] args) {
    
            User user1 = new User("001","小张",25);
            User user2 = new User("002","小王",15);
            User user3 = new User("003","小黄",19);
    
            List<User> userList = new ArrayList<>();
            userList.add(user1);
            userList.add(user2);
            userList.add(user3);
    
    
            System.out.println("==========");
            Map<Integer, List<User>> integerListMap = groupBy(userList, User::getAge);
            System.out.println(integerListMap);
    
        }
    
    
        public static <K, E> Map<K, List<E>> groupBy(Collection<E> list, Function<E, K> function) {
            return list.stream().collect(groupingBy(function));
        }

直接使用Java8提供的groupby函数即可，运行上面的代码，观察效果

![watermark_type_d3F5LXplbmhlaQ_shadow_50_text_Q1NETiBA6YCG6aOO6aOe57-U55qE5bCP5Y-U_size_20_color_FFFFFF_t_70_g_se_x_16 2][]

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