Question

我有以下问题：给定这些类，

class Person {
    private String zip;
    ...
    public String getZip(){
        return zip;
    }
}

class Region {
    private List<String> zipCodes;
    ...
    public List<String> getZipCodes() {
        return zipCodes;
    }
}

使用Java 8 Stream API，如何根据Map<Person, List<Region>>是否包含Region的邮政编码获取Person？换句话说，如何通过邮政编码属于这些区域的人对区域进行分组？

我在Java 7中用老式的方式完成了它，但现在我必须迁移代码以利用Java 8中的新功能。

谢谢，

英佩特

Answer 1

我怀疑这是最干净的方式 - 我对发布的其他答案不太满意 - 会是

 persons.stream().collect(Collectors.toMap(
    person -> person,
    person -> regions.stream()
       .filter(region -> region.getZipCodes().contains(person.getZip()))
       .collect(Collectors.toList())));

Answer 2

原始答案使用元组进行不必要的映射，因此您可以看到最终解决方案。您可以删除映射，只需直接过滤regions列表：

//A Set<Region> is more appropriate, IMO
.stream()
.collect(toMap(p -> p, 
               p -> regions.stream()
                           .filter(r -> r.getZipCodes().contains(p.getZip()))
                           .collect(toSet())));

<小时/> 如果我理解得很好，你可以这样做：

import java.util.AbstractMap.SimpleEntry;
import static java.util.stream.Collectors.toMap;
import static java.util.stream.Collectors.toList;

...

List<Person> persons = ...;
List<Region> regions = ...;

Map<Person, List<Region>> map = 
    persons.stream()
           .map(p -> new SimpleEntry<>(p, regions))
           .collect(toMap(SimpleEntry::getKey, 
                          e -> e.getValue().stream()
                                           .filter(r -> r.getZipCodes().contains(e.getKey().getZip()))
                                           .collect(toList())));

从List<Person>获得Stream<Person>。然后，将每个实例映射到包含所有区域的元组<Person, List<Region>>。从那里，您使用toMap收集器在地图中收集数据，并为每个人构建一个包含该人的邮政编码的Region列表。

例如，给定输入：

List<Person> persons = Arrays.asList(new Person("A"), new Person("B"), new Person("C"));

List<Region> regions = 
     Arrays.asList(new Region(Arrays.asList("A", "B")), new Region(Arrays.asList("A")));

输出：

Person{zip='A'} => [Region{zipCodes=[A, B]}, Region{zipCodes=[A]}]
Person{zip='B'} => [Region{zipCodes=[A, B]}]
Person{zip='C'} => []

另外，我猜每个zipCodes的{{1}}可能是Region。

Answer 3

我没有对此代码进行了任何测试，但它编译所以必须正确（：eyeroll :)。

public Map<Person,List<Region>> mapPeopleToRegion(List<Person> people, List<Region> regions){
    final Map<Person,List<Region>> personToRegion = new HashMap<>();
    people.forEach(person ->
          personToRegion.put(
                person,regions.stream().filter(
                      region -> region.getZipCodes().contains(person.getZip()))
                      .collect(Collectors.toList())));
    return personToRegion;
}

Answer 4

它仍然非常丑陋，我认为通过改变你的模型来改善它会有所改进，但到目前为止我只能设法得到以下内容：

public static void main(String[] args) {
    Person[] people = {new Person("00001"), new Person("00002"), new Person("00005")};
    Region[] regions = {
            new Region("Region 1", Arrays.asList("00001", "00002", "00003")),
            new Region("Region 2", Arrays.asList("00002", "00003", "00004")),
            new Region("Region 3", Arrays.asList("00001", "00002", "00005"))
    };

    Map<Person, List<Region>> result = Stream.of(regions)
            .flatMap(region -> region.getZipCodes().stream()
                    .map(zip -> new SimpleEntry<>(zip, region)))
            .flatMap(entry -> Stream.of(people)
                    .filter(person -> person.getZip().equals(entry.getKey()))
                    .map(person -> new SimpleEntry<>(person, entry.getValue())))
            .collect(Collectors.groupingBy(Entry::getKey, Collectors.mapping(Entry::getValue, Collectors.toList())));

    result.entrySet().forEach(entry -> System.out.printf("[%s]: {%s}\n", entry.getKey(), entry.getValue()));

    //      Output:
    //      [Person: 0]: {[name: Region 1, name: Region 3]}
    //      [Person: 1]: {[name: Region 1, name: Region 2, name: Region 3]}
    //      [Person: 2]: {[name: Region 3]}
}

拥有一个包含映射并且可以键入的ZipCode类可以使事情更清晰：

public static void main(String[] args) {
        Region r1 = new Region("Region 1");
        Region r2 = new Region("Region 2");
        Region r3 = new Region("Region 3");

        ZipCode zipCode1 = new ZipCode("00001", Arrays.asList(r1, r3));
        ZipCode zipCode2 = new ZipCode("00002", Arrays.asList(r1, r2, r3));
        ZipCode zipCode3 = new ZipCode("00003", Arrays.asList());
        ZipCode zipCode4 = new ZipCode("00004", Arrays.asList());
        ZipCode zipCode5 = new ZipCode("00005", Arrays.asList(r3));

        Person[] people = {
                new Person(zipCode1),
                new Person(zipCode2),
                new Person(zipCode5)
        };

        Map<Person, List<Region>> result = Stream.of(people)
            .collect(Collectors.toMap(person -> person,
                    person -> person.getZip().getRegions()));

        result.entrySet().forEach(entry -> System.out.printf("[%s]: {%s}\n", entry.getKey(), entry.getValue()));

//      Output:
//      [Person: 0]: {[name: Region 1, name: Region 3]}
//      [Person: 1]: {[name: Region 1, name: Region 2, name: Region 3]}
//      [Person: 2]: {[name: Region 3]}
}

Answer 5

其他一些答案包含对列表进行大量线性搜索的代码。我认为Java 8 Stream解决方案不应该比经典变体慢得多。所以这是一个利用Streams而不会牺牲很多性能的解决方案。

List<Person> people = ...
List<Region> regions = ...

Map<String, List<Region>> zipToRegions =
    regions.stream().collect(
        () -> new HashMap<>(),
        (map, region) -> {
            for(String zipCode: region.getZipCodes()) {
                List<Region> list = map.get(zipCode);
                if(list == null) list = new ArrayList<>();
                list.add(region);
                map.put(zipCode, list);
            }
        },
        (m1, m2) -> m1.putAll(m2)
    );
Map<Person, List<Region>> personToRegions =
  people.stream().collect(
    Collectors.toMap(person -> person,
                     person -> zipToRegions.get(person.getZip()))
  );

如何在Java 8中按另一个元素对List的元素进行分组

5 个答案: