小编给大家分享一下Java8怎样通过Lambda处理List集合，希望大家阅读完这篇文章之后都有所收获，下面让我们一起去探讨吧！
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
计算
安全
数据库
网络和加速
企业服务
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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