过滤不同的组并获取解析为 Integer 或 Double 的 String 值的平均值

Question

我有一个如下所示的 POJO：

@Data
class Employee {
    private int id;
    private String name;
    private String salary;
    private String departmentName;
}

还有如下列表：

    List<Employee> employees = new ArrayList<>();
    Employee emp1 = new Employee(1, "Jiya Brein", "4000", "HR");
    Employee emp2 = new Employee(2, "Paul Niksui", "2000", "IT");
    Employee emp3 = new Employee(3, "Martin Theron", "5000", "HR");
    Employee emp4 = new Employee(4, "Murali Gowda", "6000", "IT");
    Employee emp5 = new Employee(5, "Jacob Arthur", "8000", "HR");
    employees.add(emp1);
    employees.add(emp2);
    employees.add(emp3);
    employees.add(emp4);
    employees.add(emp5);

现在我正在尝试按部门筛选和分组所有员工，需要获得支出最高的部门。

如果工资是 Double 类型（比如）我可以像下面那样做：

        Employee emp1 = new Employee(1, "Jiya Brein", 4000d, "HR");
        Employee emp2 = new Employee(2, "Paul Niksui", 2000d, "IT");
        Employee emp3 = new Employee(3, "Martin Theron", 5000d, "HR");
        Employee emp4 = new Employee(4, "Murali Gowda", 6000d, "IT");
        Employee emp5 = new Employee(5, "Jacob Arthur", 8000d, "HR");

        Map<String, Double> avgSalaryOfDepartments = employees.stream().collect(
                Collectors.groupingBy(Employee::getDepartmentName, Collectors.averagingDouble(Employee::getSalary)));
        Set<Entry<String, Double>> entrySet = avgSalaryOfDepartments.entrySet();
        Map.Entry<String, Double> maxSpenderDept = null;
        for (Entry<String, Double> entry : entrySet) {
            System.out.println(entry.getKey() + " : " + entry.getValue());
            if (maxSpenderDept == null || entry.getValue().compareTo(maxSpenderDept.getValue()) > 0) {
                maxSpenderDept = entry;
            }
        }
        return maxSpenderDept.toString();

我得到了预期的输出：

HR : 5666.666666666667
IT : 4000.0
The highest spending department is HR=5666.666666666667

但是如果我得到 String 形式的薪水并且必须使用 Java 8 的流链接将其解析为 Double，我不知道该怎么办？

Answer 1

你可以边解析边取平均。尝试使用这个 lambda：

emp -> Double.parseDouble(emp.getSalary())

代替此方法参考：

Employee::getSalary

完整管道：

Map<String, Double> avgSalaryOfDepartments = 
                employees.stream()
                         .collect(groupingBy(Employee::getDepartmentName, 
                                             averagingDouble(emp -> Double.parseDouble(emp.getSalary()))));

Answer 2

以防万一你想仅使用流直接查找 max avg salary 部门名称

 String deptWithMaxAvgSal = employees.stream().collect(
            Collectors.groupingBy(Employee::departmentName,
                    Collectors.averagingDouble(emp->Double.parseDouble(emp.salary())))).entrySet().stream().max(Map.Entry.comparingByValue()).get().getKey();

过滤不同的组并获取解析为 Integer 或 Double 的 String 值的平均值

Filter distinct groups and get average of String values parsing as Integer or Double

java

collections

java-8

java-stream

double