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如何在Java 8中按另一个元素对List的元素进行分组

[英]How to group elements of a List by elements of another in Java 8

I have the following problem: Given these classes, 我有以下问题:鉴于这些类,

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

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

using the Java 8 Stream API, how do I obtain a Map<Person, List<Region>> based on whether the Region contains that Person 's zip code? 使用Java 8 Stream API,如何根据Region是否包含Person的邮政编码获取Map<Person, List<Region>> In other words how do I group the regions by the people whose zip codes belong to those regions? 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?

I've done it in Java 7 the old fashioned way, but now I have to migrate the code to take advantage of the new features in Java 8. 我已经用Java 7老式地完成了它,但现在我必须迁移代码以利用Java 8中的新功能。

Thank you, 谢谢,

Impeto 英佩特

I suspect the cleanest way to do this -- I'm not quite happy with the other answers posted -- would be 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())));

The original answer does an unnecessary mapping with tuples, so you see there the final solution. 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的解决方案。 You could remove the mapping, and simply filter directly the regions list: 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())));


If I understand well, you could do something like this: 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()))); 

From the List<Person> you get a Stream<Person> . List<Person>您将获得Stream<Person> Then you map each instance to a tuple <Person, List<Region>> that contains all the regions. 然后将每个实例映射到包含所有区域的元组<Person, List<Region>> From there, you collect the data in a map with the toMap collector and, for each person, you build a List of Region that contains the zip code of that person. 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。

For example, given the input: 例如,给定输入:

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

It outputs: 它输出:

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

Also I guess the zipCodes for each Region could be a Set . 另外我猜每个RegionzipCodes都可以是Set

I have not done any testing of this code, but it compiles so it must be right (:eyeroll:). 我还没有对这段代码进行任何测试,但它编译得一定是对的(: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;
}

It's still pretty ugly, and I think it would be improved by changing how you model things a bit, but I've only managed to come up with the following so far: 它仍然非常难看,而且我认为通过改变你对它的建模方式可以改进它,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]}
}

Having a ZipCode class that contained the mapping and could be keyed on would make things cleaner: 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]}
}

Some of the other answers contain code that does a lot of linear searching through lists. 其他一些答案包含对列表进行大量线性搜索的代码。 I think the Java 8 Stream solution should not be much slower than the classical variant. 我认为Java 8 Stream解决方案不应该比经典变体慢得多。 So here is a solution that takes advantage of Streams without sacrificing much performance. 所以这是一个利用Streams而不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()))
  );

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