收集基于多个字段的ID列表

Question

我有一个personId、年龄和性别的person对象。

public class Person {
private int personId;
private int age;
private int gender; // 0 for male and 1 for female
}

List<Person> person = new Arraylist<>();
person.add(new Person(1,1,1));
person.add(new Person(2,2,0));
person.add(new Person(3,10,1));
person.add(new Person(4,11,0));
person.add(new Person(5,20,1));
person.add(new Person(6,20,1));
person.add(new Person(7,2,0));
person.add(new Person(8,20,0));
person.add(new Person(9,11,0));
person.add(new Person(10,20,1));

我想创建一个像这样的临时对象，包含年龄、性别和学生ID列表。

TempObject {
private int age;
private int gender;
private List<Integer> studentIds;
}

现在，我想创建带有年龄、性别和学生ID列表的TempObject。这个对象应该有一对年龄、性别和与年龄和性别相对应的学生ID列表。有人能帮我吗。我已经尝试过使用java8的.分组

new TempObject(1,1,[1]);
new TempObject(2,0,[2,7]);
new TempObject(10,1,[3]);
new TempObject(11,0,[4,9]);
new TempObject(20,1,[5,6,10]);
new TempObject(20,0,[8]);

Answer 1

您可以在这里观看非常好的指南

不管怎样，我希望它对你有帮助(也许有一点点(。

主要类别

import java.util.ArrayList;
import java.util.List;
import java.util.stream.Collectors;
public class mainMethod {
public static void main(String[] args) {
List<Person> persons = new ArrayList<>();
persons.add(new Person(1, 1, 1));
persons.add(new Person(2, 2, 0));
persons.add(new Person(3, 1, 1));
persons.add(new Person(4, 11, 0));
persons.add(new Person(5, 20, 1));
persons.add(new Person(6, 20, 1));
persons.add(new Person(7, 2, 0));
persons.add(new Person(8, 20, 0));
persons.add(new Person(9, 11, 0));
persons.add(new Person(10, 20, 1));
TempObjectMapper tempObjectMapper = new TempObjectMapper(persons.stream()
.collect(Collectors.groupingBy(Person::getAge, Collectors.groupingBy(Person::getGender))));
List<TempObject> tempObjects = tempObjectMapper.getObjects();
System.out.println(tempObjects.toString());
}
}

TempObjectMapper

import java.util.ArrayList;
import java.util.List;
import java.util.Map;
public class TempObjectMapper {
private Map<Integer, Map<Integer, List<Person>>> map;
public TempObjectMapper(Map<Integer, Map<Integer, List<Person>>> collect) {
this.map = collect;
}
public List<TempObject> getObjects() {
List<TempObject> list = new ArrayList<TempObject>();
this.map.forEach((key, value) -> {
int age = key;
Map<Integer,List<Person>> map1 = value;
map1.forEach((key1, value1) -> {
int gender = key1;
List<Person> person = value1;
list.add(new TempObject(age, gender, person));
});
});
return list;
}
}

TempObject

import java.util.List;
public class TempObject {

private int age;
private int gender;
private List<Person> persons;
public TempObject(int age, int gender, List<Person> persons) {
this.age = age;
this.gender = gender;
this.persons = persons;
}
@Override
public String toString() {
return String.format("TempObject: [%s,%s,%s]" , this.age, this.gender, this.persons.toString());
}
}

个人

public class Person {
private int personId;
private int age;
private int gender;  //0 for male and 1 for female
public Person(int id, int age, int gender) {
this.personId = id;
this.age = age;
this.gender = gender;
}
public int getPersonId() {
return this.personId;
}
public int getAge() {
return this.age;
}
public int getGender() {
return this.gender;
}
@Override
public String toString() {
return String.format("Person: [%s, %s, %s]", this.personId,this.age,this.gender);
}
}

你可以用这行过滤你的列表

persons.stream()              
.collect(Collectors.groupingBy(Person::getAge, Collectors.groupingBy(Person::getGender)))

Answer 2

您可以使用Collectors.toMap(keyMapper,valueMapper,mergeFunction,mapFactory)方法收集通过两个字段进行比较的重复TreeMap：年龄和性别，并将ID合并到列表中：

List<Person> persons = Arrays.asList(
new Person(1, 1, 1),
new Person(2, 2, 0),
new Person(3, 10, 1),
new Person(4, 11, 0),
new Person(5, 20, 1),
new Person(6, 20, 1),
new Person(7, 2, 0),
new Person(8, 20, 0),
new Person(9, 11, 0),
new Person(10, 20, 1));

ArrayList<TempObject> tempObjects =
new ArrayList<>(persons.stream()
// convert Person to TempObject
.map(e -> new TempObject(e.getAge(), e.getGender(),
Collections.singletonList(e.getPersonId())))
// collect a Map<TempObject, TempObject>
.collect(Collectors.toMap(
// key of the map
Function.identity(),
// value of the map
Function.identity(),
// merge function
(to1, to2) -> {
// merging two lists of ids
to1.setStudentIds(List
.of(to1.getStudentIds(), to2.getStudentIds())
.stream()
.flatMap(List::stream)
.distinct()
.collect(Collectors.toList()));
return to1;
},
// map factory - specify a comparator
// for duplicates by age and gender
() -> new TreeMap<>(Comparator
.comparing(TempObject::getGender)
.thenComparing(TempObject::getAge))))
// get a Collection
// of the values
.values());

tempObjects.forEach(System.out::println);
// age=2, gender=0, studentIds=[2, 7]
// age=11, gender=0, studentIds=[4, 9]
// age=20, gender=0, studentIds=[8]
// age=1, gender=1, studentIds=[1]
// age=10, gender=1, studentIds=[3]
// age=20, gender=1, studentIds=[5, 6, 10]

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^{另请参阅：排序列表<映射<字符串，对象>gt；基于值}