过滤不同的组，并获得解析为Integer或Double的String值的平均值

Question

我有一个POJO，如下所示：

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

下面的列表如下：

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，我正试图按部门对所有员工进行筛选和分组，并需要获得支出最高的部门。

如果工资类型是Double(比如(，我可以这样做：

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，如果我得到的薪水是String，并且必须使用Java8的流链接将其解析为Double，我不知道该怎么办？

Answer 1

您可以在求平均值的同时解析它。尝试使用此lambda:

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

代替此方法参考：

Employee::getSalary

---

完整的管道：

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

Answer 2

只是为了防止您想通过仅使用流直接找到最大平均工资部门名称

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