在Java8+中过滤map列表元素并返回map

Question

我想过滤map列表的元素，然后在Java8+中返回map

public class Test {
    List<Employee> list1 = new ArrayList<>();
    list1.add(emp1);                   // emp1's  filterEmployee return true    
    list1.add(emp2);                   // emp2's  filterEmployee return true  
    list1.add(emp3);                   // emp3's  filterEmployee return false  

    List<Employee> list2 = new ArrayList<>();
    list2.add(emp4);                    // emp4's  filterEmployee return false        
    list2.add(emp5);                    // emp5's  filterEmployee return true 
    list2.add(emp6);                    // emp6's  filterEmployee return true 

    map.put("Sales", list1);
    map.put("Tech", list2);



    public Map<Department, List<Employee>> getEmployeeByDepartment( Map<Department, List<Employee>> map)
    {
        return map1;
    }   

    static boolean filterEmployee(Employee employee)
    {
        check employee something and filter out based on some critiea
        and return true or false based on that
    }
}

在getEmployeeByDepartment方法中，对于映射中的每个条目，使用下面的filterEmployee方法在映射值的列表中筛选employee，返回的map1将包含映射，以便

<“tech”，list2.add(emp6)>//emp6的filterEmployee返回true

我在下面试过这个和那个，但没有成功

Map<Department, List<Employee>> map2 = map.entrySet().stream().filter(entry -> entry.getValue().stream().filter(Test::filterEmployee).collect(Collectors.toMap((Department)entry.getKey(), (List<Employee>)entry.getValue())));

Answer 1

类似这样的事情应该会起作用：

private static boolean shouldKeep(Employee employee) { ... }

// ...

Map<Department, List<Employee>> employees = // a map you've built

// ...

Map<Department, List<Employee>> filteredMap =
                employees.entrySet().stream().collect(
                        Collectors.toMap(
                                e -> e.getKey(),
                                e -> e.getValue().stream()
                                        .filter(employee -> shouldKeep(employee))
                                        .collect(Collectors.toList())));