I have following two classes:
Person:
public class Person {
private final Long id;
private final String address;
private final String phone;
public Person(Long id, String address, String phone) {
this.id = id;
this.address = address;
this.phone = phone;
}
public Long getId() {
return id;
}
public String getAddress() {
return address;
}
public String getPhone() {
return phone;
}
@Override
public String toString() {
return "Person [id=" + id + ", address=" + address + ", phone=" + phone + "]";
}
}
CollectivePerson:
集体人：
import java.util.HashSet;
import java.util.Set;
public class CollectivePerson {
private final Long id;
private final Set<String> addresses;
private final Set<String> phones;
public CollectivePerson(Long id) {
this.id = id;
this.addresses = new HashSet<>();
this.phones = new HashSet<>();
}
public Long getId() {
return id;
}
public Set<String> getAddresses() {
return addresses;
}
public Set<String> getPhones() {
return phones;
}
@Override
public String toString() {
return "CollectivePerson [id=" + id + ", addresses=" + addresses + ", phones=" + phones + "]";
}
}
我希望有流操作，以便实现以下目的：
- 将
Person映射到CollectivePerson - 将
Person的address和phone合并到CollectivePerson中的addresses和phones，对于所有具有相同id的Person
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Objects;
import java.util.stream.Collectors;
public class Main {
public static void main(String[] args) {
Person person1 = new Person(1L, "Address 1", "Phone 1");
Person person2 = new Person(2L, "Address 2", "Phone 2");
Person person3 = new Person(3L, "Address 3", "Phone 3");
Person person11 = new Person(1L, "Address 4", "Phone 4");
Person person21 = new Person(2L, "Address 5", "Phone 5");
Person person22 = new Person(2L, "Address 6", "Phone 6");
List<Person> persons = new ArrayList<>();
persons.add(person1);
persons.add(person11);
persons.add(person2);
persons.add(person21);
persons.add(person22);
persons.add(person3);
Map<Long, CollectivePerson> map = new HashMap<>();
List<CollectivePerson> collectivePersons = persons.stream()
.map((Person person) -> {
CollectivePerson collectivePerson = map.get(person.getId());
if (Objects.isNull(collectivePerson)) {
collectivePerson = new CollectivePerson(person.getId());
map.put(person.getId(), collectivePerson);
collectivePerson.getAddresses().add(person.getAddress());
collectivePerson.getPhones().add(person.getPhone());
return collectivePerson;
} else {
collectivePerson.getAddresses().add(person.getAddress());
collectivePerson.getPhones().add(person.getPhone());
return null;
}
})
.filter(Objects::nonNull)
.collect(Collectors.<CollectivePerson>toList());
collectivePersons.forEach(System.out::println);
}
}
它完成了工作，并输出如下内容：
CollectivePerson [id=1, addresses=[Address 1, Address 4], phones=[Phone 1, Phone 4]]
CollectivePerson [id=2, addresses=[Address 2, Address 6, Address 5], phones=[Phone 5, Phone 2, Phone 6]]
CollectivePerson [id=3, addresses=[Address 3], phones=[Phone 3]]
但我相信有一种更好的方式，流式分组可以实现相同的效果。任何指针都将是很棒的。
HashMap::new；这个任务不要求该映射是HashMap的实例... - HolgertoMap变体还需要一个供应商。话虽如此，我刚刚注意到三个参数的toMap（即没有供应商）的实现是return toMap(keyMapper, valueMapper, mergeFunction, HashMap::new);:) - EranHashMap，就像toList()总是生成一个ArrayList一样，但是这并不是保证的。如果没有要求获取这些类型的确切实例，您应该允许实现根据更改承诺的任何好处进行更改。而且正好相反，没有采用映射供应商而不需要合并函数的方法。 - Holger