java 8、lambda、Objects copying: Creating a new list of Normalized objects

Question

从 REST 服务中，我将获得作为员工列表的响应。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 >，上面的响应需要以规范化的方式处理，如下定义。

[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;
    // 50 other properties

    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;
    // 50 other properties

    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。

EDIT: 员工 class 拥有大约 50 个房产。

如何创建这种标准化形式的列表？

Answer 1

流解决方案：

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 来消除重复实例。

我会选择简单的循环：

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，并将实例映射到id。如果id是第一次遇到，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。

编辑： 由于 computeIfAbsent 由于许多属性而不受欢迎，因此您可以不添加参数构造函数并使用 setter 传递属性值。最好的选择是使用映射库，然后你也可以用 computeIfAbsent 来做。

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> employeeIdMap2 = new HashMap<>();
        for (Employee employee : employees) {
            int id = employee.getEmpId();
            EmployeeNormalized normalized = employeeIdMap2.get(id);
            if (normalized == null) {
                normalized = new EmployeeNormalized();
                normalized.setEmpId(id);
                normalized.setName(employee.getName());
                normalized.setAddresses(new ArrayList<>());
                //set other properties
                //or even better use mapping library to create normalized and transfer property values
                employeeIdMap2.put(id, normalized);
            }
            normalized.getAddresses().add(employee.getAddress());
        }
        List<EmployeeNormalized> employeeNormalizedList2 = new ArrayList<>(employeeIdMap2.values());
        employeeNormalizedList2.forEach(System.out::println);
    }
}

Answer 2

这个任务可以用 Stream API 来解决，前提是有一些包装器 class/record 来表示 key（employeeId 和 employeeName）。即使是普通的 ArrayList<Object> 也可以用于此目的，但是引入 record 因为 Java 16 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，Collectors.mapping 构建每个员工的地址列表，最后加入键（employeeId 和 employeeName) 和 EmployeeNormalized:

中地址的值列表

//
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 可以在不进行额外转换的情况下维护类型安全的键。 Record 可以用一个小包装器 class 代替，以表示具有 hashCode / equals 方法被覆盖的键，因为此 class 的实例用作中间映射中的键。

// 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());

java 8、lambda、Objects copying: Creating a new list of Normalized objects

java 8, lambda , Objects copying: Creating a new list of Normalized objects

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