Java 8 - リストをマップに変換する

Java 8 –リストをマップに変換

オブジェクトのListMapに変換する方法、および重複したキーを処理する方法を示すJava8の例はほとんどありません。

Hosting.java

package com.example.java8

public class Hosting {

    private int Id;
    private String name;
    private long websites;

    public Hosting(int id, String name, long websites) {
        Id = id;
        this.name = name;
        this.websites = websites;
    }

    //getters, setters and toString()
}

1. マップリスト– Collectors.toMap()

Hostingオブジェクトのリストを作成し、Collectors.toMapを使用してそれをマップに変換します。

TestListMap.java

package com.example.java8

import java.util.ArrayList;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;

public class TestListMap {

    public static void main(String[] args) {

        List list = new ArrayList<>();
        list.add(new Hosting(1, "liquidweb.com", 80000));
        list.add(new Hosting(2, "linode.com", 90000));
        list.add(new Hosting(3, "digitalocean.com", 120000));
        list.add(new Hosting(4, "aws.amazon.com", 200000));
        list.add(new Hosting(5, "example.com", 1));

        // key = id, value - websites
        Map result1 = list.stream().collect(
                Collectors.toMap(Hosting::getId, Hosting::getName));

        System.out.println("Result 1 : " + result1);

        // key = name, value - websites
        Map result2 = list.stream().collect(
                Collectors.toMap(Hosting::getName, Hosting::getWebsites));

        System.out.println("Result 2 : " + result2);

        // Same with result1, just different syntax
        // key = id, value = name
        Map result3 = list.stream().collect(
                Collectors.toMap(x -> x.getId(), x -> x.getName()));

        System.out.println("Result 3 : " + result3);
    }
}

出力

Result 1 : {1=liquidweb.com, 2=linode.com, 3=digitalocean.com, 4=aws.amazon.com, 5=example.com}
Result 2 : {liquidweb.com=80000, example.com=1, digitalocean.com=120000, aws.amazon.com=200000, linode.com=90000}
Result 3 : {1=liquidweb.com, 2=linode.com, 3=digitalocean.com, 4=aws.amazon.com, 5=example.com}

2. マップリスト-重複キー!

2.1 Run below code, and duplicated key errors will be thrown!

TestDuplicatedKey.java

package com.example.java8;

import java.util.ArrayList;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;

public class TestDuplicatedKey {

    public static void main(String[] args) {

        List list = new ArrayList<>();
        list.add(new Hosting(1, "liquidweb.com", 80000));
        list.add(new Hosting(2, "linode.com", 90000));
        list.add(new Hosting(3, "digitalocean.com", 120000));
        list.add(new Hosting(4, "aws.amazon.com", 200000));
        list.add(new Hosting(5, "example.com", 1));

        list.add(new Hosting(6, "linode.com", 100000)); // new line

        // key = name, value - websites , but the key 'linode' is duplicated!?
        Map result1 = list.stream().collect(
                Collectors.toMap(Hosting::getName, Hosting::getWebsites));

        System.out.println("Result 1 : " + result1);

    }
}

出力-以下のエラーメッセージは少し誤解を招く可能性があり、キーの値の代わりに「linode」を表示する必要があります。

Exception in thread "main" java.lang.IllegalStateException: Duplicate key 90000
    at java.util.stream.Collectors.lambda$throwingMerger$0(Collectors.java:133)
    at java.util.HashMap.merge(HashMap.java:1245)
    //...

2.2 To solve the duplicated key issue above, pass in the third mergeFunction argument like this :

    Map result1 = list.stream().collect(
                Collectors.toMap(Hosting::getName, Hosting::getWebsites,
                        (oldValue, newValue) -> oldValue
                )
        );

出力

Result 1 : {..., aws.amazon.com=200000, linode.com=90000}

Note
(oldValue, newValue) -> oldValue =⇒キーが重複している場合、oldKeyとnewKeyのどちらを使用しますか?

3.3 Try newValue

    Map result1 = list.stream().collect(
                Collectors.toMap(Hosting::getName, Hosting::getWebsites,
                        (oldValue, newValue) -> newvalue
                )
        );

出力

Result 1 : {..., aws.amazon.com=200000, linode.com=100000}

3. リストからマップ–並べ替えと収集

TestSortCollect.java

package com.example.java8;

import java.util.*;
import java.util.stream.Collectors;

public class TestSortCollect {

    public static void main(String[] args) {

        List list = new ArrayList<>();
        list.add(new Hosting(1, "liquidweb.com", 80000));
        list.add(new Hosting(2, "linode.com", 90000));
        list.add(new Hosting(3, "digitalocean.com", 120000));
        list.add(new Hosting(4, "aws.amazon.com", 200000));
        list.add(new Hosting(5, "example.com", 1));
        list.add(new Hosting(6, "linode.com", 100000));

        //example 1
        Map result1 = list.stream()
                .sorted(Comparator.comparingLong(Hosting::getWebsites).reversed())
                .collect(
                        Collectors.toMap(
                                Hosting::getName, Hosting::getWebsites, // key = name, value = websites
                                (oldValue, newValue) -> oldValue,       // if same key, take the old key
                                LinkedHashMap::new                      // returns a LinkedHashMap, keep order
                        ));

        System.out.println("Result 1 : " + result1);

    }
}

出力

Result 1 : {aws.amazon.com=200000, digitalocean.com=120000, linode.com=100000, liquidweb.com=80000, example.com=1}

P.S In above example, the stream is sorted before collect, so the “linode.com=100000” became the ‘oldValue’.