Java8-将List转成Map

几个Java 8的例子展示怎样将一个对象的集合（List）放入一个Map中，并且展示怎样处理多个重复keys的问题。

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()
}

1. List to Map – Collectors.toMap()

创建一个 Hosting 对象集合, 并且用 Collectors.toMap 去将它转换放入一个 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);
    }
}

输出结果

1
2
3

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}

2. List转Map 重复key问题

以下代码会抛出异常

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);
 
    }
}

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)
	//...

为了解决上面重复key的问题，通过增加第三个参数解决:

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

输出

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

(oldValue, newValue) -> oldValue ==> 如果key是重复的，你选择oldKey or newKey?

如果是用新的值

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

输出如下

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

3. 排序后转换

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);
 
    }
}

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

P.S 在上面的例子中， stream 在collect 之前已经被排序, 所以 “linode.com=100000”变为 ‘oldValue’.

1. List to Map – Collectors.toMap()

2. List转Map 重复key问题

3. 排序后转换

References