java8新特性：利用Lambda处理List集合

时间：2020-01-21 16:25:19 阅读：124 评论：0 收藏：0 [点我收藏+]

标签：map sum new lambda bar 结果 ima lists ati

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：

技术图片

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：

技术图片

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：

技术图片

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：

技术图片

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：

技术图片

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：

技术图片