java中Collectors.toMap() 方法使用技巧

「爱情、让人受尽委屈。」 2023-09-27 16:33 120阅读 0赞

## java中Collectors.toMap() 方法使用技巧 ##

> Collectors.toMap() 方法是把[List转Map][List_Map]的操作。

***代码***

public static void main(String[] args) {
        
    		List<Employee> employeeList = Arrays.asList(
    				new Employee(101, "张三", 18, 9999.99),
    				new Employee(102, "李四", 59, 6666.66),
    				new Employee(103, "王五", 28, 3333.33),
    				new Employee(104, "赵六", 8, 7777.77)
    		);
    		/**
    		 * id作为Map的key,name作为value的集合
    		 * */
    		Map<Integer, String> collect1 = employeeList.stream().collect(Collectors.toMap(Employee::getId, Employee::getName));
    		System.out.println(collect1);//{101=张三, 102=李四, 103=王五, 104=赵六}
    		/**
    		 * id作为map的集合，Employee对象作为Map的value
    		 */
    		Map<Integer, Employee> collect2 = employeeList.stream().collect(Collectors.toMap(Employee::getId, t -> t));
    		System.out.println(collect2);//{101=Employee(id=101, name=张三, age=18, salary=9999.99), 102=Employee(id=102, name=李四, age=59, salary=6666.66), 103=Employee(id=103, name=王五, age=28, salary=3333.33), 104=Employee(id=104, name=赵六, age=8, salary=7777.77)}
    		/**
    		 * id作为map的集合，Employee对象作为Map的value
    				*/
    		Map<Integer, Employee> collect3 = employeeList.stream().collect(Collectors.toMap(Employee::getId, Function.identity()));
    		System.out.println(collect3);//{101=Employee(id=101, name=张三, age=18, salary=9999.99), 102=Employee(id=102, name=李四, age=59, salary=6666.66), 103=Employee(id=103, name=王五, age=28, salary=3333.33), 104=Employee(id=104, name=赵六, age=8, salary=7777.77)}
    	}

***在上面代码中，map的ID不能重复，如果重复就会报错，下面我们引入个重载方法。***

/**
    		 * id作为Map的key,name作为value的集合。 如果ID相等，则返回第一个id的name
    		 */
    		Map<Integer, String> collect1 = employeeList.stream().collect(Collectors.toMap(Employee::getId, Employee::getName, (n1, n2) -> n1));
    		System.out.println(collect1);
    		/**
    		 * id作为Map的key,name作为value的集合。 如果ID相等，则返回第一个id的Employee
    		 */
    		Map<Integer, Employee> collect2 = employeeList.stream().collect(Collectors.toMap(Employee::getId, t -> t, (n1, n2) -> n1));
    		System.out.println(collect2);
    		/**
    		 * id作为Map的key,name作为value的集合。 如果ID相等，则返回第一个id的Employee
    		 */
    		Map<Integer, Employee> collect3 = employeeList.stream().collect(Collectors.toMap(Employee::getId, Function.identity(), (n1, n2) -> n1));
    		System.out.println(collect3);

[List_Map]: https://so.csdn.net/so/search?q=List%E8%BD%ACMap&spm=1001.2101.3001.7020