如何使用Java Stream将User列表转换为含SSNs、IDs双键的列表值Map

我有一个User(id, ssn, name)类型的列表，需要把它转换成包含两个键（"SSNs"、"IDs"）的Map，其中"SSNs"对应SSN字符串列表，"IDs"对应ID整数列表。

示例输入 #

List.of(new User(1, "ssn1", "user1"), new User(2, "ssn2", "user2"), new User(3, "ssn3", "user3"))

期望输出 #

Map with 2 keys("SSNs","IDs") and list as value
"SSNs" -  List("ssn1","ssn2","ssn3")
"IDs" -  List(1,2,3)

当前代码 #

public class Test {

    public static void main(String[] args) {

        List<User> users = List.of(new User(1, "ssn1", "user1"), new User(2, "ssn2", "user2"),
                new User(3, "ssn3", "user3"));

        Map<String, List<Object>> userMap = users.stream().collect(Collectors.groupingBy(User::getSsn));
    
    }

}

class User {
    public int getId() {
        return id;
    }

    public String getSsn() {
        return ssn;
    }

    public String getName() {
        return name;
    }

    int id;
    String ssn;
    String name;

    public User(int id, String ssn, String name) {
        this.id = id;
        this.ssn = ssn;
        this.name = name;
    }
}

**更新：**已将List<User>更新为List<Object>。

解决方案 #

当前代码使用Collectors.groupingBy(User::getSsn)是按SSN值分组，得到的是键为SSN、值为对应User列表的Map，和需求不符。以下是几种符合需求的实现方式：

方法一：手动遍历构建Map（兼容所有Java版本） #

最直观的方式，遍历一次列表填充两个集合后放入Map：

public static void main(String[] args) {
    List<User> users = List.of(new User(1, "ssn1", "user1"), new User(2, "ssn2", "user2"),
            new User(3, "ssn3", "user3"));

    Map<String, List<Object>> userMap = new HashMap<>();
    List<Object> ids = new ArrayList<>();
    List<Object> ssns = new ArrayList<>();

    for (User user : users) {
        ids.add(user.getId());
        ssns.add(user.getSsn());
    }

    userMap.put("IDs", ids);
    userMap.put("SSNs", ssns);
    
    // 验证输出
    System.out.println(userMap);
}

方法二：两次Stream收集（Java 8+） #

写法简洁，通过两次Stream分别收集ID和SSN列表：

public static void main(String[] args) {
    List<User> users = List.of(new User(1, "ssn1", "user1"), new User(2, "ssn2", "user2"),
            new User(3, "ssn3", "user3"));

    Map<String, List<Object>> userMap = new HashMap<>();
    userMap.put("IDs", users.stream().map(User::getId).collect(Collectors.toList()));
    userMap.put("SSNs", users.stream().map(User::getSsn).collect(Collectors.toList()));
    
    System.out.println(userMap);
}

方法三：使用Collectors.teeing单次遍历（Java 12+） #

通过teeing收集器在一次Stream遍历中同时完成两种收集，效率更高：

public static void main(String[] args) {
    List<User> users = List.of(new User(1, "ssn1", "user1"), new User(2, "ssn2", "user2"),
            new User(3, "ssn3", "user3"));

    Map<String, List<Object>> userMap = users.stream()
            .collect(Collectors.teeing(
                    Collectors.mapping(User::getId, Collectors.toList()),
                    Collectors.mapping(User::getSsn, Collectors.toList()),
                    (ids, ssns) -> {
                        Map<String, List<Object>> map = new HashMap<>();
                        map.put("IDs", ids);
                        map.put("SSNs", ssns);
                        return map;
                    }
            ));
    
    System.out.println(userMap);
}

内容的提问来源于stack exchange，提问作者Jericho