Question

我想将Java 8流用于以下实现。基本上我想解析一个列表并形成另一个不同对象的列表。

输入 - 人物pojos列表，输出 - PersonInfo pojos列表

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

    Person max = new Person();
    max.setName("Max");
    max.setAge(10);
    max.addAddress(new Address("Street1", "City1"));
    max.addAddress(new Address("Street2", "City2"));

    Person peter = new Person();
    peter.setName("Peter");
    peter.setAge(20);
    peter.addAddress(new Address("Street1", "City1"));
    peter.addAddress(new Address("Street2", "City2"));

    persons.add(max);
    persons.add(peter);

    System.out.println("Input: " + persons);

    List<PersonInfo> personInfos = new ArrayList<PersonInfo>();
    PersonInfo personInfo = null;
    for (Person person : persons) {
        for (Address addr : person.getCurrAndPrevAddrs()) {
            personInfo = new PersonInfo();
            personInfo.setName(person.getName());
            personInfo.setAge(person.getAge());
            personInfo.setAddrs(addr);              
            personInfos.add(personInfo);
        }
    }

    System.out.println("Output: " + personInfos.toString());

示例输出：输入：[最多10 [Street1 City1，Street2 City2]

，Peter 20 [Street1 City1，Street2 City2]]

输出：[Max 10 Street1 City1

，Max 10 Street2 City2

，Peter 20 Street1 City1

，Peter 20 Street2 City2]

Answer 1

List<PersonInfo> personInfos = persons.stream().flatMap(person -> person.getCurrAndPrevAddrs().stream().map(addr -> {
        PersonInfo personInfo = new PersonInfo();
        personInfo.setName(person.getName());
        personInfo.setAge(person.getAge());
        personInfo.setAddrs(addr);              
        return personInfo;
})).collect(Collectors.toList());

Answer 2

以下是您尝试执行的操作的简化版本。我为我的测试简化了Pojos。

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

Person person1 = new Person("person1");
Person person2 = new Person("person2");

persons.add(person1);
persons.add(person2);

List<PersonInfo> personInfos = new ArrayList<PersonInfo>();
persons.stream().forEach(person -> {    
  person.getCurrAndPrevAddrs().stream().forEach(address -> {
    PersonInfo personInfo = new PersonInfo("personInfo");
    personInfo.setAddress(address);
    personInfos.add(personInfo);
  });
});

System.out.println("Output: " + personInfos.toString());

Answer 3

（1）提高可读性和（2）简化代码维护的一种方法是向toPersonInfo类添加Person方法，如下所示：

public class Person {
    //getters, setters, etc.
    public List<PersonInfo> toPersonInfos() {
        List<PersonInfo> result = new ArrayList<>();
        for (Address addr : getCurrentAndPrevAddrs()) {
            PersonInfo pi = new PersonInfo();
            pi.setName(this::getName);
            pi.setAge(this::getAge);
            pi.setAddrs(this::addr);
            result.add(pi);
        }
        return result;
    }
}

这样，如果您更改了PersonInfo或Person课程，则只需更改一种方法，它就在您的Person课程中。

它还极大地简化了流操作：

personInfos = persons.stream().flatMap(p -> p.toPersonInfos.stream()).collect(Collectors.toList());

请注意，这不一定会返回有序列表。

如何用Java 8流替换嵌套循环

3 个答案: