Java的8个多分组和减少

我有一种情况，我有一个类型的对象的列表，我需要从它创建另一种类型的另一个对象的列表。

下面是代码：我有员工和需要的列表，从第一个列表创建EmployeeInfo的名单。需要注意的是员工有一个帐户属性和EmployeeInfo有帐户的列表。在这种情况下，同样的员工可以有多个账户，所以在employeeinfo列表中产生，每个信息对象将有帐户列表。下面是如何我都做到了：

public class Employee {

private final int dept;

private final String name;

private final String city;

private final String account;

public Employee(int dept, String name, String city, String account) {
    this.dept = dept;
    this.name = name;
    this.city = city;
    this.account = account;
}

public int getDept() {
    return dept;
}

public String getName() {
    return name;
}

public String getCity() {
    return city;
}

public String getAccount() {
    return account;
}

}

public class EmployeeInfo {

private final String name;

private final List<String> accounts;

public EmployeeInfo(String name, List<String> accounts) {
    this.name = name;
    this.accounts = accounts;
}

public String getName() {
    return name;
}

public List<String> getAccounts() {
    return accounts;
}

public EmployeeInfo addToList(EmployeeInfo employeeInfo) {
    List<String> l = new ArrayList<>();
    l.addAll(this.getAccounts());
    l.addAll(employeeInfo.getAccounts());
    return new EmployeeInfo(employeeInfo.name, l);
}

}

测试类：

public static void main(String[] args){
    List<Employee> employees = new ArrayList<>();
    //tradeId, secPool, datasetid, restricValue
    employees.add(new Employee(1, "Mary", "Boston", "A1"));
    employees.add(new Employee(1, "Mary", "Boston", "A2"));
    employees.add(new Employee(1, "Alex", "NYC", ""));
    employees.add(new Employee(2, "Peter", "DC", ""));
    employees.add(new Employee(1, "Sophia", "DC", "A4"));

    TestEmployeeGrouping testEmployeeGrouping = new TestEmployeeGrouping();

    Map<Integer, List<EmployeeInfo>> result = new HashMap<>();
    Map<Integer, Map<String, List<Employee>>> map =  employees.stream().collect(groupingBy(Employee::getDept, groupingBy(testEmployeeGrouping::createKey)));
    map.forEach((integer, stringListMap) -> {
        List<EmployeeInfo> employeeInfos = createInfo(stringListMap);
        result.put(integer, employeeInfos);
    });


}

private static List<EmployeeInfo> createInfo(Map<String,List<Employee>> stringListMap) {
    List<EmployeeInfo> employeeInfos = new ArrayList<>();
    stringListMap.forEach((s, employees) -> {
        List<String> accounts = employees.stream().map(Employee::getAccount).collect(Collectors.toList());
        employeeInfos.add(new EmployeeInfo(employees.get(0).getName(), accounts));
    });
    return employeeInfos;
}


private String createKey(Employee employee) {
    return employee.getDept() + employee.getName();
}

虽然上面的一块工作正常，最后给我employeeinfo的列表，通过部门分组，每个都有其帐户列表，我想这样做在爱茉莉泛函的方式，如：

employees.stream().collect(groupingBy(Employee::getDept, groupingBy(testEmployeeGrouping::createKey, reducing(EmployeeInfo::addToList))));

上述线抛出错误：不兼容的类型：T是无法转换为雇员。是否有人可以帮助我弄清楚如何解决这个问题？

谢谢！