通过在两个其他列表中匹配条目的新Java对象列表

YCF的解决方案应该有效,但它是一个O(N ^二 )解决方案。通过将一个列表转换为从ID映射到对象的映射,然后在另一个列表上迭代并从映射中获取匹配值,可以实现O(N)解决方案:

Map<Integer, Person> personMap = 
    persons.stream().collect(Collectors.toMap(Person::getId, Function.identity());

List<PersonWithMetadata> result = 
    metadata.stream()
            .map(m -> new PersonWithMetadata(personMap.get(m.getId()), m)
            .collect(Collectors.toList());

在示例数据中,列表具有匹配顺序的匹配对象。如果这个假设对实际问题也是正确的,那么解决方案就必须更容易——您可以将 索引 并从列表中获取相应的值:

List<PersonWithMetadata> result = 
    IntStream.reange(0, persons.size())
             .map(i -> new PersonWithMetadata(persons.get(i), metadata.get(i))
             .collect(Collectors.toList());

您可以这样尝试:

List<PersonWithMetadata> personsAndMEtadata = persons.stream()
        .map(p -> {
                    //search for the meta data based on the person id
                    Metadata meta = metadata.stream()
                            .filter(m -> m.getId() == p.getId())
                            .findFirst()
                            .get();
                    // then create a PersonWithMetadata object based on Person and metadata
                    return new PersonWithMetadata(
                            p.getId(), p.getFirstName(), p.getLastName(),
                            meta.getDate(), meta.getCity(), meta.getJob()
                    );

                }
        ).collect(Collectors.toList());

关于此行:

Metadata meta = metadata.stream().filter(m -> m.getId() == p.getId()).findFirst().get();

我假设您有一个ID为Person的元数据,否则您将得到 NullPointerException .

我相信你要找的是 zip 很遗憾,API中省略了函数。

这个 protonpack library 提供它,这将允许您压缩tuple,然后将其映射到新结构。

StreamUtils.zip(persons, metadata, (person, metadata) -> ... )

下面的示例构建 Map 属于 Metadata 使用ID作为键的对象。这将有助于提高性能,因为不需要迭代 元数据 每个的列表 Person 在 List

代码

public static void main(String[] args) {
    List<Person> persons = Arrays.asList(
            new Person(1, "Mike", "Canada"),
            new Person(2, "Jill", "England"),
            new Person(3, "Will", "Whales"),
            new Person(4, "Mary", "Spain"));


    List<Metadata> metadataList = Arrays.asList(
            new Metadata(1, "2000-01-01", "Naturalized", "Bachelor's of Arts"),
            new Metadata(2, "2001-01-01", "ExPat", "Masters of Chemestry"),
            new Metadata(3, "2017-05-01", "Citizen", "Buiness Management"),
            new Metadata(4, "2018-04-16", "Work Visa", "Nursing"));

    //Iterate over metadataList once and create map based on ID as key
    Map<Integer, List<Metadata>> metadataMap = metadataList.stream()
            .collect(Collectors.groupingBy(Metadata::getId));

    //Iterate over personList and fetch metadata from Map to build PersonWithMetadata
    List<PersonWithMetadata> personWithMetadataList = persons.stream().map(person -> {
        List<Metadata> metadata = metadataMap.get(person.id);
        if (metadata.isEmpty()) {
            //TODO: Handle scenario for no metadata for person
        }

        //TODO: Build PersonWithMetadata

        return new PersonWithMetadata();

    }).collect(Collectors.toList());

}