带有单独收集的java流过滤器
我有一份名单;每个人都有人类型,即10、11、12
我想制作一个子列表,每个子列表都有人类型
对于Streams,我通过3次迭代实现了这一点
Can group&;参与1次迭代
人。爪哇
public class Person {
String personType;
String name;
public Person(String personType, String name) {
this.personType = personType;
this.name = name;
}
public String getPersonType() {
return personType;
}
public void setPersonType(String personType) {
this.personType = personType;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
@Override
public String toString() {
return "Person{" +
"personType='" + personType + '\'' +
", name='" + name + '\'' +
'}';
}
}
司机。爪哇
import java.util.ArrayList;
import java.util.List;
import java.util.stream.Collectors;
public class Driver {
public static void main(String[] args) {
List<Person> persons = populateList();
List<Person> persons1 = persons.stream().filter( p -> p.getPersonType().equals("10")).collect(Collectors.toList());
List<Person> persons2 = persons.stream().filter( p -> p.getPersonType().equals("10")).collect(Collectors.toList());
List<Person> persons3 = persons.stream().filter( p -> p.getPersonType().equals("10")).collect(Collectors.toList());
}
public static List<Person> populateList(){
Person person01 = new Person("10","Tony");
Person person02 = new Person("11","Steve");
Person person03 = new Person("12","Banner");
Person person04 = new Person("10","Thor");
Person person05 = new Person("11","Natasha");
Person person06 = new Person("12","Loki");
Person person07 = new Person("10","Peter");
Person person08 = new Person("11","HawkEye");
Person person09 = new Person("12","Falcon");
Person person10 = new Person("10","Jarvis");
List<Person> persons = new ArrayList<>();
persons.add(person01);
persons.add(person02);
persons.add(person03);
persons.add(person04);
persons.add(person05);
persons.add(person06);
persons.add(person07);
persons.add(person08);
persons.add(person09);
persons.add(person10);
return persons;
}
}
# 1 楼答案
你可以使用分组方式