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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