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：

以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持我们。