1

我有一个关于 Java 8 流的具体问题。即使听起来不太复杂，我也找不到任何接近解决我的问题的方法。

我想将 aList<Person>转换为Map<Company, List<Person>>. 该公司是一个枚举。现在Map应该按Company（到目前为止，我一切正常。在我想通过一个属性过滤所有的之后： 的名称List<Person>。我只想让里面有最长名字的人。示例：MapPerson

这是我到目前为止得到的：

import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.stream.Collectors;

public class PersonMap {

    public static void main(String[] args) {
        Person person1 = new Person("Pete", Company.APPLE);
        Person person2 = new Person("Joe", Company.APPLE);
        Person person3 = new Person("Jack", Company.APPLE);
        Person person4 = new Person("Robert", Company.MICROSOFT);
        Person person5 = new Person("Cate", Company.MICROSOFT);
        Person person6 = new Person("Abi", Company.MICROSOFT);

        List<Person> list = new LinkedList<>();
        list.add(person1);
        list.add(person2);
        list.add(person3);
        list.add(person4);
        list.add(person5);
        list.add(person6);
    }

    public Map<Company, List<Person>> groupPersonByCompanyAndFilterByName (List <Person> list){
        Map<Company, List<Person>> outputMap =
                list.stream().collect(Collectors.groupingBy(Person::getCompany))
                .entrySet().stream()
                .filter(p -> p.getValue().stream().toString().length() !=
                p.getValue().stream().map(Person::getName).mapToInt(String::length).max().getAsInt())
                .collect(Collectors.toMap(Entry::getKey, Entry::getValue));

        return outputMap;
    }
}

过滤功能在这里不起作用。我想要的输出是：

{MICROSOFT=["person4"], APPLE=[person1, person3]}

有没有人有好的知识可以帮助我！

4

4 回答 4

2

这是一种使用发球的方法

public Map<Company, List<Person>> groupPersonByCompanyAndFilterByName (List <Person> list){
    return list.stream().collect(groupingBy(Person::getCompany,
           teeing(
                  mapping(Person::getName, maxBy(Comparator.comparing(String::length))),
                  toList(),
                  (a, b) -> b.stream().filter(x -> x.getName().length() == a.get().length())
                                      .collect(toList())

           )
    ));

}
于 2020-01-04T15:26:25.870 回答
0

为了获得更好的性能，您不应在单个流链中执行此操作。

试试这个：

List<Person> list = new LinkedList<>();
list.add(new Person("Pete", Company.APPLE));
list.add(new Person("Joe", Company.APPLE));
list.add(new Person("Jack", Company.APPLE));
list.add(new Person("Robert", Company.MICROSOFT));
list.add(new Person("Cate", Company.MICROSOFT));
list.add(new Person("Abi", Company.MICROSOFT));

Map<Company, List<Person>> map = list.stream().collect(Collectors.groupingBy(Person::getCompany));
map.values().forEach(persons -> {
    int maxLen = persons.stream().mapToInt(p -> p.getName().length()).max().getAsInt();
    persons.removeIf(p -> p.getName().length() != maxLen);
});

System.out.println(map);

输出

{APPLE=[Pete, Jack], MICROSOFT=[Robert]}

对于上面的代码，我使用了以下额外的类：

enum Company {
    APPLE, MICROSOFT
}

class Person {
    private final String name;
    private final Company company;
    public Person(String name, Company company) {
        this.name = name;
        this.company = company;
    }
    public String getName() {
        return this.name;
    }
    public Company getCompany() {
        return this.company;
    }
    @Override
    public String toString() {
        return this.name;
    }
}
于 2020-01-04T15:01:56.947 回答
0

您可以对多个字段Company和长度进行分组name

Map<Company, Map<Integer, List<Person>>> map = list.stream()
            .collect(Collectors.groupingBy(Person::getCompany, Collectors.groupingBy(p -> p.getName().length())));

并且Entries从外部地图key和value具有最大长度（名称）的内部地图创建，这是内部地图key

Map<Company, List<Person>> resultMap = map.entrySet().stream()
            .map(e -> new AbstractMap.SimpleEntry<>(e.getKey(),
                    e.getValue().entrySet().stream()
                            .max(Comparator.comparing(Map.Entry<Integer, List<Person>>::getKey)).get().getValue()))
            .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));

您也可以将这两个操作组合成单个流

于 2020-01-04T16:02:55.660 回答
-2 
     Person person1 = new Person("Pete", "APPLE");
            Person person2 = new Person("Joe", "APPLE");
            Person person3 = new Person("Jack", "APPLE");
            Person person4 = new Person("Robert", "MICROSOFT");
            Person person5 = new Person("Cate", "MICROSOFT");
            Person person6 = new Person("Abi", "MICROSOFT");

            List<Person> list = new ArrayList<>();

            list.add(person1);
            list.add(person2);
            list.add(person3);
            list.add(person4);
            list.add(person5);
            list.add(person6);

  //Adapt your structure and with this function you can group  
            Map<String, List<Person>> map = list.stream().collect(Collectors.groupingBy(person -> person.getCompany()));

            for (String name: map.keySet()){
                List<Person> persons = map.get(name);
                System.out.println("[" + name + "]");
                persons.forEach(s -> System.out.println(s.getName()));
            }

Out Log: [APPLE] Pete Joe Jack [MICROSOFT] Robert Cate Abi

于 2020-01-04T14:52:21.763 回答