Java 8映射值的对象列表

Question

public class User{
   String name;
   String age;
   String org;
}

 m.put("names", List of names,);
 m.put("age", List of age);
 m.put("org", List of org);

List<String> names = userList.stream().map(User::getName).collect(Collectors.toList());
List<String> age= userList.stream().map(User::getAge).collect(Collectors.toList());
List<String> org= userList.stream().map(User::getName).collect(Collectors.toList());

Answer 1

我相信这样的事情应该会奏效：

Map<String,List<String>> map =
    userList.stream()
            .flatMap(user -> {
                Map<String,String> um = new HashMap<>();
                um.put("names",user.getName());
                um.put("age",user.getAge());
                um.put("org",user.getOrg());
                return um.entrySet().stream();
            }) // produces a Stream<Map.Entry<String,String>>
            .collect(Collectors.groupingBy(Map.Entry::getKey,
                                           Collectors.mapping(Map.Entry::getValue,
                                                              Collectors.toList())));

它将每个用户转换为映射 （包含由必需的键索引的3个必需属性），然后将所有用户映射的条目按其键分组。

编辑：

Map<String,List<String>> map =
    userList.stream()
            .flatMap (user -> Stream.of (new SimpleEntry<>("names",user.getName()),
                                         new SimpleEntry<>("age",user.getAge()),
                                         new SimpleEntry<>("org",user.getOrg())))
            .collect(Collectors.groupingBy(Map.Entry::getKey,
                                           Collectors.mapping(Map.Entry::getValue,
                                                              Collectors.toList())));