Java8如何通過Lambda處理List集合

更新時間：2020年01月22日 12:00:20 作者：weizhxa

這篇文章主要介紹了java8如何通過Lambda處理List集合,文中通過示例代碼介紹的非常詳細，對大家的學習或者工作具有一定的參考學習價值,需要的朋友可以參考下

Java 8新增的Lambda表達式，我們可以用簡潔高效的代碼來處理List。

1、遍歷

public static void main(String[] args) {
  List<User> userList = Lists.newArrayList();

  User user1 = new User(1L, "張三", 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());
  });

 }

運行結(jié)果：

2、list轉(zhuǎn)為Map

public static void main(String[] args) {
  List<User> userList = Lists.newArrayList();//存放user對象集合

  User user1 = new User(1L, "張三", 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為key，轉(zhuǎn)為Map
  Map<Long,User> userMap = userList.stream().collect(Collectors.toMap(User::getId, a -> a,(k1, k2)->k1));
  System.out.println(userMap);

 }

運行結(jié)果：

3、將List分組：List里面的對象元素，以某個屬性來分組

public static void main(String[] args) {
  List<User> userList = Lists.newArrayList();//存放user對象集合

  User user1 = new User(1L, "張三", 24);
  User user2 = new User(2L, "李四", 27);
  User user3 = new User(3L, "王五", 21);
  User user4 = new User(4L, "張三", 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);

  //根據(jù)name來將userList分組
  Map<String, List<User>> groupBy = userList.stream().collect(Collectors.groupingBy(User::getName));
  System.out.println(groupBy);

 }

運行結(jié)果：

4、過濾：從集合中過濾出來符合條件的元素

public static void main(String[] args) {
  List<User> userList = Lists.newArrayList();//存放user對象集合

  User user1 = new User(1L, "張三", 24);
  User user2 = new User(2L, "李四", 27);
  User user3 = new User(3L, "王五", 21);
  User user4 = new User(4L, "張三", 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);


  //取出名字為張三的用戶
  List<User> filterList = userList.stream().filter(user -> user.getName().equals("張三")).collect(Collectors.toList());
  filterList.stream().forEach(user ->{
   System.out.println(user.getName());
  });

 }

運行結(jié)果：

5、求和：將集合中的數(shù)據(jù)按照某個屬性求和

public static void main(String[] args) {
  List<User> userList = Lists.newArrayList();//存放user對象集合

  User user1 = new User(1L, "張三", 24);
  User user2 = new User(2L, "李四", 27);
  User user3 = new User(3L, "王五", 21);
  User user4 = new User(4L, "張三", 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);


  //取出名字為張三的用戶
  int totalAge = userList.stream().mapToInt(User::getAge).sum();
  System.out.println("和：" + totalAge);

 }

運行結(jié)果：

6、從List轉(zhuǎn)為Map，key與value 一一對應

public static void main(String[] args) {
  List<User> userList = Lists.newArrayList();

  User user1 = new User(1L, "張三", 24);
  User user2 = new User(2L, "李四", 27);
  User user3 = new User(3L, "王五", 21);
  User user4 = new User(4L, "張三", 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));

 }

運行結(jié)果：