Java8如何通过Lambda处理List集合

这篇文章主要介绍了java8如何通过Lambda处理List集合,文中通过示例代码介绍的非常详细，对大家的学习或者工作具有一定的参考学习价值,需要的朋友可以参考下

Java 8新增的Lambda表达式，我们可以用简洁高效的代码来处理List。

1、遍历

public static void main(String[] args) {
  List<User> userList = Lists.newArrayList();

  User user1 = new User(1L, "张三", 24);
  User user2 = new User(2L, "李四", 27);
  User user3 = new User(3L, "王五", 21);

  userList.add(user1);
  userList.add(user2);
  userList.add(user3);

  userList.stream().forEach(user ->{
   System.out.println(user.getName());
  });

 }

运行结果：

2、list转为Map

public static void main(String[] args) {
  List<User> userList = Lists.newArrayList();//存放user对象集合

  User user1 = new User(1L, "张三", 24);
  User user2 = new User(2L, "李四", 27);
  User user3 = new User(3L, "王五", 21);

  userList.add(user1);
  userList.add(user2);
  userList.add(user3);

  //ID为key，转为Map
  Map<Long,User> userMap = userList.stream().collect(Collectors.toMap(User::getId, a -> a,(k1, k2)->k1));
  System.out.println(userMap);

 }

运行结果：

3、将List分组：List里面的对象元素，以某个属性来分组

public static void main(String[] args) {
  List<User> userList = Lists.newArrayList();//存放user对象集合

  User user1 = new User(1L, "张三", 24);
  User user2 = new User(2L, "李四", 27);
  User user3 = new User(3L, "王五", 21);
  User user4 = new User(4L, "张三", 22);
  User user5 = new User(5L, "李四", 20);
  User user6 = new User(6L, "王五", 28);

  userList.add(user1);
  userList.add(user2);
  userList.add(user3);
  userList.add(user4);
  userList.add(user5);
  userList.add(user6);

  //根据name来将userList分组
  Map<String, List<User>> groupBy = userList.stream().collect(Collectors.groupingBy(User::getName));
  System.out.println(groupBy);

 }

运行结果：

4、过滤：从集合中过滤出来符合条件的元素

public static void main(String[] args) {
  List<User> userList = Lists.newArrayList();//存放user对象集合

  User user1 = new User(1L, "张三", 24);
  User user2 = new User(2L, "李四", 27);
  User user3 = new User(3L, "王五", 21);
  User user4 = new User(4L, "张三", 22);
  User user5 = new User(5L, "李四", 20);
  User user6 = new User(6L, "王五", 28);

  userList.add(user1);
  userList.add(user2);
  userList.add(user3);
  userList.add(user4);
  userList.add(user5);
  userList.add(user6);


  //取出名字为张三的用户
  List<User> filterList = userList.stream().filter(user -> user.getName().equals("张三")).collect(Collectors.toList());
  filterList.stream().forEach(user ->{
   System.out.println(user.getName());
  });

 }

运行结果：

5、求和：将集合中的数据按照某个属性求和

public static void main(String[] args) {
  List<User> userList = Lists.newArrayList();//存放user对象集合

  User user1 = new User(1L, "张三", 24);
  User user2 = new User(2L, "李四", 27);
  User user3 = new User(3L, "王五", 21);
  User user4 = new User(4L, "张三", 22);
  User user5 = new User(5L, "李四", 20);
  User user6 = new User(6L, "王五", 28);

  userList.add(user1);
  userList.add(user2);
  userList.add(user3);
  userList.add(user4);
  userList.add(user5);
  userList.add(user6);


  //取出名字为张三的用户
  int totalAge = userList.stream().mapToInt(User::getAge).sum();
  System.out.println("和：" + totalAge);

 }

运行结果：

6、从List转为Map，key与value 一 一对应

public static void main(String[] args) {
  List<User> userList = Lists.newArrayList();

  User user1 = new User(1L, "张三", 24);
  User user2 = new User(2L, "李四", 27);
  User user3 = new User(3L, "王五", 21);
  User user4 = new User(4L, "张三", 22);
  User user5 = new User(5L, "李四", 20);
  User user6 = new User(6L, "王五", 28);

  userList.add(user1);
  userList.add(user2);
  userList.add(user3);
  userList.add(user4);
  userList.add(user5);
  userList.add(user6);

  Map<Long/*Id*/,User> userMap = userList.stream().collect(Collectors.toMap(User::getId, user -> user));
  System.out.println("toMap：" + JSONArray.toJSONString(userMap));

 }

运行结果：

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