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
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