[英]java 8, lambda , Objects copying: Creating a new list of Normalized objects
从 REST 服务中,我将得到员工列表的响应。 其中可能包含以下定义的同一员工的多个地址。
[Employee{empId=1, name='Emp1', address='Emp1 Address1'},
Employee{empId=1, name='Emp1', address='Emp1 Address 2'},
Employee{empId=2, name='Emp2', address='Emp2 Address 1'}]
通过创建另一个列表,即List<EmployeeNormalized >
,需要以标准化方式处理上述响应,如下所述。
[EmployeeNormalized{empId=1, name='Emp1',
addresses=[Emp1 Address1, Emp1 Address 2]},
EmployeeNormalized{empId=2, name='Emp2', addresses=[Emp2 Address 1]}]
代码片段:
class Employee {
private int empId;
private String name;
private String address;
public Employee(int empId, String name, String address) {
this.empId = empId;
this.name = name;
this.address = address;
}
// Setters and Getters
}
class EmployeeNormalized {
private int empId;
private String name;
private List<String> addresses;
public EmployeeNormalized(int empId, String name, List<String> address) {
this.empId = empId;
this.name = name;
this.addresses = address;
}
// Setters and Getters
}
List<EmployeeNormalized >
必须包含唯一的员工对象,并且EmployeeNormalized
class 中的List<String>
应该包含该员工的所有地址。
如何创建这种标准化形式的列表?
Stream 解决方案:
public class Normalize {
public static void main(String[] args) {
List<Employee> employees = new ArrayList<>();
employees.add(new Employee(1, "Emp1", "Address 1"));
employees.add(new Employee(1, "Emp1", "Address 2"));
employees.add(new Employee(2, "Emp2", "Address 3"));
List<EmployeeNormalized> employeeNormalizedList = employees.stream()
.map(new Function<Employee, EmployeeNormalized>() {
private final Map<Integer, EmployeeNormalized> employeeIdMap = new HashMap<>();
@Override
public EmployeeNormalized apply(Employee employee) {
EmployeeNormalized normalized = this.employeeIdMap.computeIfAbsent(employee.getEmpId(),
key -> new EmployeeNormalized(employee.getEmpId(), employee.getName(), new ArrayList<>()));
normalized.getAddresses().add(employee.getAddress());
return normalized;
}
})
.distinct()
.collect(Collectors.toList());
employeeNormalizedList.forEach(System.out::println);
}
}
相当复杂,需要调用 distinct 来消除重复的实例。
我会 go 用于简单的循环:
public class Normalize {
public static void main(String[] args) {
List<Employee> employees = new ArrayList<>();
employees.add(new Employee(1, "Emp1", "Address 1"));
employees.add(new Employee(1, "Emp1", "Address 2"));
employees.add(new Employee(2, "Emp2", "Address 3"));
Map<Integer, EmployeeNormalized> employeeIdMap = new HashMap<>();
for (Employee employee : employees) {
EmployeeNormalized normalized = employeeIdMap.computeIfAbsent(employee.getEmpId(), key -> new EmployeeNormalized(employee.getEmpId(), employee.getName(), new ArrayList<>()));
normalized.getAddresses().add(employee.getAddress());
}
List<EmployeeNormalized> employeeNormalizedList = new ArrayList<>(employeeIdMap.values());
employeeNormalizedList.forEach(System.out::println);
}
}
基本上,这两种解决方案都使用员工 ID 作为唯一标识符,并使用 map 实例作为 ID。 如果第一次遇到id,则创建实例并添加地址,如果已经有该id的实例,则获取实例并添加地址。
这个任务可以用 Stream API 来解决,前提是有一些包装类/记录来表示一个键(employeeId 和employeeName)。 即使是普通的ArrayList<Object>
也可以用于此目的,但自 Java 16 以来引入record
,最好使用它们。
解决方案本身非常简单:使用Collectors.groupingBy
创建一个键,使用Collectors.mapping
构建每个员工的地址列表,最后在EmployeeNormalized
中加入键( employeeId
和employeeName
)和地址值列表:
//
List<Employee> employees = .... ; // build employee list
List<EmployeeNormalized> normEmployees = employees
.stream()
.collect(Collectors.groupingBy(
emp -> Arrays.asList(emp.getEmpId(), emp.getName()),
LinkedHashMap::new, // maintain insertion order
Collectors.mapping(Employee::getAddress, Collectors.toList())
)) // Map<List<Object>, List<String>>
.entrySet()
.stream()
.map(e -> new EmployeeNormalized(
((Integer) e.getKey().get(0)).intValue(), // empId
(String) e.getKey().get(1), // name
e.getValue()
))
.collect(Collectors.toList());
使用record
允许维护类型安全的键而无需额外的转换。 可以用一个小的包装器 class 替换记录,以表示一个带有hashCode
/ equals
方法的键,因为此 class 的实例用作中间 map 中的键。
// Java 16+
record EmpKey(int empId, String name) {}
List<EmployeeNormalized> normEmployees = employees
.stream()
.collect(Collectors.groupingBy(
emp -> new EmpKey(emp.getEmpId(), emp.getName()),
LinkedHashMap::new, // maintain insertion order
Collectors.mapping(Employee::getAddress, Collectors.toList())
)) // Map<List<Object>, List<String>>
.entrySet()
.stream()
.map(e -> new EmployeeNormalized(
e.getKey().empId(), e.getKey().name(), e.getValue()
))
.collect(Collectors.toList());
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.