Java 8 – Convert List to Map（将 List 转换为 Map）

短命女 2022-06-05 00:25 186阅读 0赞

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。

Hosting.java

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.

TestListMap.java

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); } }

Output

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 to Map – Duplicated Key! （List转Map 重复key问题） ##

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

TestDuplicatedKey.java

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：下面这个错误有的误导，它应该显示 “linode” 代替key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)
    	//...

2.2  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 ) );

Output

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

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

3.3 Try newValue

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

Output

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

## 3. List to Map – Sort & Collect ##

TestSortCollect.java

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); } }

Output

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’.

## References ##

1.  [Java 8 Collectors JavaDoc][]
2.  [Java 8 – How to sort a Map][Java 8 _ How to sort a Map]
3.  [Java 8 Lambda : Comparator example][Java 8 Lambda _ Comparator example]

[Java 8 Collectors JavaDoc]: https://docs.oracle.com/javase/8/docs/api/java/util/stream/Collectors.html
[Java 8 _ How to sort a Map]: http://www.mkyong.com/java8/java-8-how-to-sort-a-map/
[Java 8 Lambda _ Comparator example]: https://www.mkyong.com/java8/java-8-lambda-comparator-example/