[JAVA] Aggregate the number of people every 10 years from List <Person>

Table of Contents ⇒ Java Algorithm Library-Artery-Sample


package jp.avaj.lib.algo;

import java.util.ArrayList;
import java.util.List;

import jp.avaj.lib.def.ArCharDef;
import jp.avaj.lib.def.ArMatch;
import jp.avaj.lib.def.ArRepeat;
import jp.avaj.lib.test.L;

 *  List<Person>Aggregate the number of people every 10 years from
 *-Use the following methods of ArList.
 *  Integer[] histogram(Collection<T0> collection,ArCreator<T0,T1> creator,ArValidator<T1>[] validators)
 *-Use creator to extract data to be aggregated from object T0 of List.
 *・ In this sample, the age is extracted from the Person object..
 *· Use validators to stratify extracted ages.
 *・ Although it is an array, it should return true every 10 years..
public class Q01_10 {
  public static void main(String[] args) throws Exception {
    List<Person> personList = createPersonList();
    //ArCreator ⇒ Extract age from Person
    ArCreator<Person,Integer> creator = new ArCreator<Person,Integer>() {
      public Integer convert(Person obj) throws Exception {
        return obj.getAge();
    //ArValidator for determining age
    ArValidatorIntRange[] validators = new ArValidatorIntRange[] {
      new ArValidatorIntRange(10,20,ArMatch.YES), // 10(Including)~20(Not included),ArMatch.YES,NO reverses the judgment
      new ArValidatorIntRange(20,30,ArMatch.YES),
      new ArValidatorIntRange(30,40,ArMatch.YES),
      new ArValidatorIntRange(40,50,ArMatch.YES),
      new ArValidatorIntRange(50,60,ArMatch.YES),

    Integer[] result;
    //Create a histogram
    result = ArList.histogram(personList,creator, validators);

    L.p("Another solution");
    //Another solution of ArCreator.Use ArCreatorByName to get the value from the field name as a generic ArCreator.
    creator = new ArCreatorByName<Person,Integer>("age");
    //Another solution of ArValidator
    //Creating ArValidator one by one like the first solution is tedious.
    //It can also be easily generated as follows.Note, null can be specified at both ends ⇒ infinite range
    validators = ArValidatorUtil.createArValidatorIntRangeArray(new Integer[]{10,20,30,40,50,60});
    result = ArList.histogram(personList,creator, validators);

  private static List<Person> createPersonList() {
    //Class for generating a name ⇒ Name is a single uppercase letter
    //ArRandomFromArray fetches the value randomly from the array, ArRepeat specifies whether to fetch the same value.
    ArRandomFromArray<String> names = new ArRandomFromArray<String>(ArCharDef.upperStr,ArRepeat.NO);
    //Class for generating age ⇒ 10(Including)~60(Not included)Randomly generate an Integer
    ArRandomFromInteger ages = new ArRandomFromInteger(10,60);

    List<Person> list = new ArrayList<Person>();
    for (int i=0; i<20; i++) {
      Person person = new Person(names.get(),ages.get());
      L.p(ArObj.toString(person)); //By doing this, the contents will be displayed even if toString is not implemented..
    return list;

  static class Person {
    public Person(String name,int age) {
      this.name = name;
      this.age = age;
    private String name;
    private int age;
    public String getName() {
      return name;
    public void setName(String name) {
      this.name = name;
    public int getAge() {
      return age;
    public void setAge(int age) {
      this.age = age;

The result is as follows


[age=31, name=S]
[age=26, name=F]
[age=42, name=K]
[age=43, name=Y]
[age=22, name=M]
[age=34, name=N]
[age=11, name=E]
[age=36, name=Q]
[age=13, name=U]
[age=41, name=X]
[age=39, name=Z]
[age=47, name=O]
[age=47, name=P]
[age=59, name=H]
[age=52, name=A]
[age=17, name=D]
[age=35, name=C]
[age=47, name=W]
[age=55, name=J]
[age=11, name=R]
[10-20) -> 4
[20-30) -> 2
[30-40) -> 5
[40-50) -> 6
[50-60) -> 3
Another solution
[10-20) -> 4
[20-30) -> 2
[30-40) -> 5
[40-50) -> 6
[50-60) -> 3

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