Java8 新特性 Stream流操作List集合 (二)

本是古典何须时尚 2022-11-22 12:57 209阅读 0赞

lambda表达式的list去重和list转map

public static List<Employee> createEmployee() {
            List<Employee> employeeList = new ArrayList<Employee>();
            Employee employee1 = new Employee();
            Employee employee2 = new Employee();
            Employee employee3 = new Employee();
    
            employee1.setEmployeeId("1");
            employee1.setEmployeeName("张三");
            employee1.setEmployeeSex("男");
    
            employee2.setEmployeeId("2");
            employee2.setEmployeeName("王五");
            employee2.setEmployeeSex("男");
    
            employee3.setEmployeeId("3");
            employee3.setEmployeeName("李四");
            employee3.setEmployeeSex("男");
    
            employeeList.add(employee1);
            employeeList.add(employee2);
            employeeList.add(employee3);
    
            return employeeList;
        }
    
        public static List<Employee> createEmployee2() {
            List<Employee> employeeList = new ArrayList<Employee>();
            Employee employee1 = new Employee();
            Employee employee2 = new Employee();
            Employee employee3 = new Employee();
            Employee employee4 = new Employee();
    
            employee1.setEmployeeId("1");
            employee1.setEmployeeName("张三");
            employee1.setEmployeeSex("男");
    
            employee2.setEmployeeId("1");
            employee2.setEmployeeName("王五");
            employee2.setEmployeeSex("男");
    
            employee1.setEmployeeId("1");
            employee1.setEmployeeName("田七");
            employee1.setEmployeeSex("男");
    
            employee3.setEmployeeId("3");
            employee3.setEmployeeName("李四");
            employee3.setEmployeeSex("男");
    
            employeeList.add(employee1);
            employeeList.add(employee2);
            employeeList.add(employee3);
    
            return employeeList;
        }
    
        public static List<String> createString() {
            List<String> list = new ArrayList<String>();
            list.add("1");
            list.add("2");
            list.add("3");
            list.add("4");
            list.add("5");
            list.add("1");
            list.add("2");
            return list;
        }
    
        public static void list() {
            // 去重list
            List<String> listString = createString();
            System.out.println(listString.stream().distinct().collect(Collectors.toList()));
    
            System.out.println("******************************************************************");
    
            // list 转 map map的value 就是 Employee 对象
            List<Employee> employeeList = createEmployee();
            Map<String, Employee> map = employeeList.stream().collect(Collectors.toMap(Employee::getEmployeeId, employee -> employee));
            map.forEach((k, v) -> {
                System.out.println("key: " + k + "    value: " + v);
            });
    
            System.out.println("******************************************************************");
    
            // list 转 map  涉及key 重复
            // 第一种写法  (k1, k2) -> k1) 的意思是保留 k1 存在重复不传第三个处理参数会报错 Duplicate key
            List<Employee> employeeList2 = createEmployee2();
            Map<String, String> map1 = employeeList2.stream().collect(Collectors.toMap(e -> e.getEmployeeId(), e -> e.getEmployeeName(), (k1, k2) -> k1));
            System.out.println(map1);
    
            System.out.println("******************************************************************");
    
            // 第二种写法
            Map map2 = employeeList2.stream().collect(HashMap::new, (map3, e) -> map3.put(e.getEmployeeId(), e.getEmployeeName()), Map::putAll);
            System.out.println(map2);
    
            System.out.println("******************************************************************");
    
            // 处理遍历中存在null 的情况  value 为 null 会报空指针
            // 异常重现
    //        Map<String, String> map3 = employeeList.stream().collect(Collectors.toMap(e -> e.getEmployeeId(), e -> e.getEmployeeBirthDay()));
    //        System.out.println(map3);
    
            // 解决异常
            Map<String, String> map3 = employeeList.stream().collect(Collectors.toMap(e -> e.getEmployeeId(), e -> Optional.ofNullable(e.getEmployeeBirthDay()).orElse("没有数据")));
            System.out.println(map3);
    
            System.out.println("******************************************************************");
    
            //  当 key 全部为 null 时 而 value 唯一就会报错出现Duplicate key value的情况
            // 异常重现
    //        Map<String, String> map4 = employeeList2.stream().collect(Collectors.toMap(Employee::getEmployeeBirthDay, Employee::getEmployeeName));
    //        System.out.println(map4);
    
            // 解决异常 指定出现重复的key之后保留的字段 (解决方法很多,中心围绕在如何解决掉重复的问题，一般都是覆盖)
            Map<String, String> map4 = employeeList2.stream().collect(Collectors.toMap(Employee::getEmployeeBirthDay, Employee::getEmployeeName, (k1, k2) -> k1));
            System.out.println(map4);
    
        }
    
        public static void main(String[] args) {
            list();
        }

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