Convert List to Map

Convert List to Map

几个java 8例子，展示了如何将List对象转化为Map，以及如何处理重复的key

package com.mkyong.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()
}

List to Map ----Collectors.toMap()

创建一个Hosting对象列表，然后使用Collectors.toMap，将该对象列表转换为Map

package com.mkyong.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<Hosting> 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, "mkyong.com", 1));

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

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

        // key = name, value - websites
        Map<String, Long> 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<Integer, String> result3 = list.stream().collect(
                Collectors.toMap(x -> x.getId(), x -> x.getName()));

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

输出

Result 1 : {1=liquidweb.com, 2=linode.com, 3=digitalocean.com, 4=aws.amazon.com, 5=mkyong.com}
Result 2 : {liquidweb.com=80000, mkyong.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=mkyong.com}

List to Map -----Duplicated key!

运行下面代码，将抛出重复key的错误

package com.mkyong.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<Hosting> 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, "mkyong.com", 1));
        
        list.add(new Hosting(6, "linode.com", 100000)); // new line

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

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

    }
}

输出，下面的错误信息有点误导，它应该展示linode，而不是键的值

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)
    //...

为了上面重复key的问题，我们可以像下面这样引入第三个归并函数

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

输出

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

(oldValue, newValue) -> oldValue ==>如果key重复，你是想要新值还是旧值

用新值

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

输出

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

List to Map -----Sort & Collect

package com.mkyong.java8;

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

public class TestSortCollect {

    public static void main(String[] args) {

        List<Hosting> 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, "mkyong.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);

    }
}

输出

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