这篇文章主要介绍了如何在Java8中将List<T>转为Map<String,T>，此处通过实例代码给大家介绍的非常详细，对大家的学习或工作具有一定的参考价值，需要的朋友可以参考下：
Java是什么Java是一门面向对象编程语言，可以编写桌面应用程序、Web应用程序、分布式系统和嵌入式系统应用程序。
将 List 转为 Map<String, T>public class AnswerApp { public static void main(String[] args) throws Exception { List<String> names = Lists.newArrayList("Answer", "AnswerAIL", "AI"); Map<String, Integer> map = names.stream().collect(Collectors.toMap(v -> v, v -> 1)); System.out.println(map); } }程序运行输出
{Answer=1, AnswerAIL=1, AI=1}将 List 转为 Map<K, V>public static void main(String[] args) throws Exception { List<User> users = new ArrayList<>(); for (int i = 0; i < 3; i++) { users.add(new User("answer" + i, new Random().nextInt(100))); } System.out.println(JSON.toJSONString(users)); System.out.println(); Map<String, Integer> map = users.stream().collect(Collectors.toMap(User::getName, User::getAge)); System.out.println(map); }程序运行输出
[{"age":78,"name":"answer0"},{"age":89,"name":"answer1"},{"age":72,"name":"answer2"}] {answer2=72, answer1=89, answer0=78}将 List 转为 Map<String, T>实现方式1
public class AnswerApp { public static void main(String[] args) throws Exception { List<User> users = new ArrayList<>(); for (int i = 0; i < 3; i++) { // 改为此代码, 转map时会报错 Duplicate key User // users.add(new User("answer", new Random().nextInt(100))); users.add(new User("answer" + i, new Random().nextInt(100))); } System.out.println(JSON.toJSONString(users)); System.out.println(); Map<String, User> map = users.stream().collect(Collectors.toMap(User::getName, Function.identity())); System.out.println(JSON.toJSONString(map)); } }该方式如果 map 的 key(如上述例子的 User::getName 的值) 重复， 会抛错java.lang.IllegalStateException: Duplicate key User
程序运行输出
[{"age":22,"name":"answer0"},{"age":79,"name":"answer1"},{"age":81,"name":"answer2"}] {"answer2":{"age":81,"name":"answer2"},"answer1":{"age":79,"name":"answer1"},"answer0":{"age":22,"name":"answer0"}}实现方式2
public class AnswerApp { public static void main(String[] args) throws Exception { List<User> users = new ArrayList<>(); for (int i = 0; i < 3; i++) { users.add(new User("answer", new Random().nextInt(100))); } System.out.println(JSON.toJSONString(users)); System.out.println(); // 如果 key 重复, 则根据 冲突方法 ·(key1, key2) -> key2· 判断. 解释: key1 key2 冲突时 取 key2 Map<String, User> map = users.stream().collect(Collectors.toMap(User::getName, Function.identity(), (key1, key2) -> key2)); System.out.println(JSON.toJSONString(map)); } }程序运行输出
[{"age":24,"name":"answer"},{"age":89,"name":"answer"},{"age":68,"name":"answer"}] {"answer":{"age":68,"name":"answer"}}如果改为 (key1, key2) -> key1 则输出 {"answer":{"age":24,"name":"answer"}}
User 实体
@Data @NoArgsConstructor @AllArgsConstructor public class User { private Long id; private String name; private Integer age; public User(String name) { this.name = name; } public User(String name, Integer age) { this.name = name; this.age = age; } }补充：java8中使用Lambda表达式将list中实体类的两个字段转Map
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