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 +
'}';
}
}
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);
}
}
