Java 8 lambda for selecting top salary employee fo

问题:

class Employee {
    public string department;
    public int salary;
}

List<Employee> allEmployees = ...

I need to have a list that will have only 1 top salary employee for each department. allEmployees is the source list.

回答1:

You can do that with a grouping collector:

Map<String, Employee> topEmployees =
    allEmployees.stream()
                .collect(groupingBy(
                    e -> e.department,
                    collectingAndThen(maxBy(comparingInt(e -> e.salary)), Optional::get) 
                ));

with the static imports

import static java.util.Comparator.comparingInt;
import static java.util.stream.Collectors.collectingAndThen;
import static java.util.stream.Collectors.groupingBy;
import static java.util.stream.Collectors.maxBy;

This code creates a Stream of all the employees and groups them with their department with the help of Collectors.groupingBy. For all the values classified to the same key, we need to keep only the employee with the maximum salary, so we collect them with Collectors.maxBy and the comparator compares the salary with Comparator.comparingInt. Since maxBy returns an Optional<Employee> (to handle the case where there the list is empty), we wrap it with a call to Collectors.collectingAndThen with a finisher that just returns the employee: we know in this case that the optional won't be empty.

回答2:

Alternative solution:

Map<String, Employee> topEmployees =
    allEmployees.stream()
            .collect(Collectors.toMap(
                e -> e.department,
                e -> e,
                BinaryOperator.maxBy(Comparator.comparingInt(e -> e.salary)) 
            ));

When we encounter the first employee from the department, we add a new entry to the Map. When another employee is found, one with higher salary is kept. This way you don't need to meddle with optionals.