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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本文标题: java8如何通过Lambda处理List集合
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