Java 8 将 list 转为 map

引言

创建一个对象类

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 Hosting 对象的列表，并使用 Collectors.toMap 将其转换为 Map。

public static void testOne(){
    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("result1：" + result1);

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

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

    // key = id, value = name 另一种写法
    Map<Integer, String> result3 = list.stream()
        .collect(Collectors.toMap(x -> x.getId(), x -> x.getName()));

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

2.List to Map – Duplicated Key

重复的 key 抛出异常。

private static void testTwo() {
    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

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

    System.out.println("result1：" + result1);
  }

输出——下面的错误消息有点误导人，它应该显示“ 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)
    //...

要解决上面重复的关键问题，传入第三个 mergeFunction 参数，如下所示:

private static void testTwo() {

    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

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

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

    System.out.println("result1：" + result1);
  }

输出：

result1：{liquidweb.com=80000, mkyong.com=1, digitalocean.com=120000, aws.amazon.com=200000, linode.com=90000}

使用新值：

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

输出：

result1：{liquidweb.com=80000, mkyong.com=1, digitalocean.com=120000, aws.amazon.com=200000, linode.com=100000}

3.List to Map – Sort & Collect

先排序再收集。

private static void testThree() {
    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));

    // use oldValue
    Map<String, Long> result1 = list.stream().sorted(Comparator.comparingLong(Hosting::getWebsites).reversed())
        .collect(Collectors.toMap(Hosting::getName,Hosting::getWebsites,(oldValue, newValue) -> oldValue,
            LinkedHashMap::new));

    System.out.println("result1：" + result1);
  }

输出：

result1：{aws.amazon.com=200000, digitalocean.com=120000, linode.com=100000, liquidweb.com=80000, mkyong.com=1}

在上面的例子中，流是在收集之前排序的，所以“ linode. com 100000”变成了“ oldValue”。

源码见：java-8-demo

系列文章详见：Java 8 教程