Java8将List转为Map

1、实体

 public class Hosting {

     private int id;

     private String name;

     private long websites;

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

     public int getId() {
         return id;
     }

     public void setId(int id) {
         this.id = id;
     }

     public String getName() {
         return name;
     }

     public void setName(String name) {
         this.name = name;
     }

     public long getWebsites() {
         return websites;
     }

     public void setWebsites(long websites) {
         this.websites = websites;
     }

     @Override
     public String toString() {
         return "Hosting{" +
                 "id=" + id +
                 ", name='" + name + '\'' +
                 ", websites=" + websites +
                 '}';
     }
 }

2、将List转为Map

 public class List2Map {

     public static void main(String[] args) {
         List<Hosting> hostings = new ArrayList<>();
         hostings.add(new Hosting(1, "liquidweb.com", 80000));
         hostings.add(new Hosting(2, "linode.com", 90000));
         hostings.add(new Hosting(3, "digitalocean.com", 120000));
         hostings.add(new Hosting(4, "aws.amazon.com", 200000));
         hostings.add(new Hosting(5, "mkyong.com", 1));

         // key = id, value = websites
         Map<Integer, String> id2Name = hostings.stream()
                 .collect(Collectors.toMap(Hosting::getId, Hosting::getName));
         System.out.println("id2Name: " + id2Name);

         // key = name, value = websites
         Map<String, Long> name2Websites = hostings.stream()
                 .collect(Collectors.toMap(Hosting::getName, Hosting::getWebsites));
         System.out.println("name2Websites: " + name2Websites);

         // key = id, value = websites
         Map<Integer, String> id2NamDifferent = hostings.stream()
                 .collect(Collectors.toMap(h -> h.getId(), h -> h.getName()));
         System.out.println("id2NamDifferent: " + id2NamDifferent);

     }
 }

3、将List转为Map(重复key的情况)

public class List2MapDuplicatedKey {

    public static void main(String[] args) {
        List<Hosting> hostings = new ArrayList<>();
        hostings.add(new Hosting(1, "liquidweb.com", 80000));
        hostings.add(new Hosting(2, "linode.com", 90000));
        hostings.add(new Hosting(3, "digitalocean.com", 120000));
        hostings.add(new Hosting(4, "aws.amazon.com", 200000));
        hostings.add(new Hosting(5, "mkyong.com", 1));

        hostings.add(new Hosting(6, "linode.com", 100000)); // 重复的key

        // key = name, vaule = websites
        Map<String, Long> name2Websites = hostings.stream()
                .collect(Collectors.toMap(Hosting::getName, Hosting::getWebsites));
        System.out.println("name2Websites: " + name2Websites);

    }
}

在上面一段代码中，"linbode.com"做为key被add两次，那么在转为map过程会发生什么?如下：

如何解决重复key的情况？只需要在16行加入如下处理即可：

 public class List2MapDuplicatedKey {

     public static void main(String[] args) {
         List<Hosting> hostings = new ArrayList<>();
         hostings.add(new Hosting(1, "liquidweb.com", 80000));
         hostings.add(new Hosting(2, "linode.com", 90000));
         hostings.add(new Hosting(3, "digitalocean.com", 120000));
         hostings.add(new Hosting(4, "aws.amazon.com", 200000));
         hostings.add(new Hosting(5, "mkyong.com", 1));

         hostings.add(new Hosting(6, "linode.com", 100000)); // 重复的key

         // key = name, vaule = websites
         Map<String, Long> name2Websites = hostings.stream()
                 .collect(Collectors.toMap(Hosting::getName, Hosting::getWebsites,
                         (oldValue, newValue) -> newValue));
         System.out.println("name2Websites: " + name2Websites);

     }
 }

4、将List转为Map并排序

 public class List2MapWithSort {

     public static void main(String[] args) {
         List<Hosting> hostings = new ArrayList<>();
         hostings.add(new Hosting(1, "liquidweb.com", 80000));
         hostings.add(new Hosting(2, "linode.com", 90000));
         hostings.add(new Hosting(3, "digitalocean.com", 120000));
         hostings.add(new Hosting(4, "aws.amazon.com", 200000));
         hostings.add(new Hosting(5, "mkyong.com", 1));
         hostings.add(new Hosting(6, "linode.com", 100000));

         // key = name, vaule = websites
         Map<String, Long> name2Websites = hostings.stream()
                 .sorted(Comparator.comparing(Hosting::getWebsites).reversed())
                 .collect(Collectors.toMap(Hosting::getName, Hosting::getWebsites,
                         (oldValue, newValue) -> newValue,   // 如果有相同的key,使用新key
                         LinkedHashMap::new));   // 返回ListedHashMap,保持有序

         System.out.println("name2Websites: " + name2Websites);
     }
 }