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java8新特性:利用Lambda处理List集合
阅读量:4542 次
发布时间:2019-06-08

本文共 4128 字,大约阅读时间需要 13 分钟。

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
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
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
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
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
> 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
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
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
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
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
userMap = userList.stream().collect(Collectors.toMap(User::getId, user -> user)); System.out.println("toMap:" + JSONArray.toJSONString(userMap)); }

运行结果:

 

转载于:https://www.cnblogs.com/JoeyWong/p/9600109.html

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