java - Java 8多级分组和归约

我有一个场景，其中有一个类型的对象列表，我需要从中创建另一个类型的另一个对象的列表。

下面是代码:我有一个雇员列表，需要从第一个列表创建EmployeeInfo列表。请注意，Employee具有一个account属性，而EmployeeInfo具有一个帐户列表。在这种情况下，同一个员工可以拥有多个帐户，因此在生成的employeeinfo列表中，每个info对象将具有一个帐户列表。这是我的方法:

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);
}

}

测试类别:

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();
}

尽管上面的方法工作正常，但最终给了我一个employeeinfo的列表，按部门将其分组，每个帐户都有其帐户列表，但我想用一种更有趣的方式做到这一点，例如:

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

上面的行引发错误:不兼容的类型:T不能转换为Employee。有人可以帮我找出解决方法吗？

谢谢!

最佳答案

您是否有任何理由将name中的EmployeeInfo设置为final？如果您可以更改此解决方案将起作用

将这两个方法添加到EmployeeInfo

public void setName(String name) {
    this.name = name;
}
public void AddAccount(String account) {
    this.accounts.add(account);
}

然后你可以做到这一点

Collector<Employee, EmployeeInfo, EmployeeInfo> empToInfo = Collector.of(
     () -> new EmployeeInfo("", new ArrayList<String>()),
            (info, e) -> {
                info.AddAccount(e.getAccount());
                info.setName(e.getName());
                },
     (p1,p2) -> p1.addToList(p2));

Collector<Employee, ?, Collection<EmployeeInfo>> byName = collectingAndThen(groupingBy(Employee::getName, empToInfo),
                  (Map<String, EmployeeInfo> finisher) -> {return finisher.values();});

Map<Integer, Collection<EmployeeInfo>> r2 = employees.stream().collect(groupingBy(Employee::getDept, byName));

如果您想保持EmployeeInfo不可变，则可以使用reduce而不是collection，它就像这样

Map<Integer, Collection<EmployeeInfo>> result2 = employees.stream().collect(groupingBy(Employee::getDept,
             collectingAndThen(groupingBy(Employee::getName, reducing(new EmployeeInfo("", new ArrayList<String>()),
                                                                      empl3 -> new EmployeeInfo(empl3.getName(),Arrays.asList(empl3.getAccount())),
                                                                      (inf1, inf2) -> inf1.addToList(inf2))),
                                finisher -> finisher.values())));