java - 使用 Lambda 对列表进行排序和分组

Question

让我们假设我有一个员工列表:

private static List<Employee> list = new ArrayList<Employee>();
    static {
        list.add(new Employee("Joe", 100000, 1980));
        list.add(new Employee("Tim", 50000, 1982));
        list.add(new Employee("Mike", 90000, 1970));
        list.add(new Employee("Rick", 50000, 1955));
        list.add(new Employee("Andy", 60000, 1966));
        list.add(new Employee("Tim", 10000, 1995));
        list.add(new Employee("Tony", 130000, 1991));
        list.add(new Employee("Timmy", 150000, 1988));
        list.add(new Employee("Rich", 50000, 1980));
        list.add(new Employee("Andrew", 160000, 1970));
        list.add(new Employee("Ton", 150000, 1958));
        list.add(new Employee("Jose", 40000, 1970));
        list.add(new Employee("Timothy", 50000, 1996));
        list.add(new Employee("Ricardo", 50000, 1988));
        list.add(new Employee("Gemasio", 60000, 1971));
        list.add(new Employee("Mike", 80000, 1992));
    }

现在我想要的是生成执行一些过滤的列表，例如:工资 > x 和工资 < y并根据员工姓名排序，如果多个员工姓名相同但薪水不同但现在使用他们的薪水，则再次排序

到目前为止我所做的是:

System.out.println(
                list.stream()
                    .filter(e -> e.getSalary() > 55000)
                    .filter(e -> e.getSalary() < 120000)
                    .sorted(Comparator.comparing(Employee::getName))
                    .collect(Collectors.groupingBy(Employee::getName))
        );

例如:仅对名称进行排序后我得到::

<name: Andy salary: 60000 year of birth: 1966>, 
    <name: Gemasio salary: 60000 year of birth: 1971>, 
    <name: Joe salary: 100000 year of birth: 1980>, 
    <name: Mike salary: 90000 year of birth: 1970>, 
    <name: Mike salary: 80000 year of birth: 1992>

这里有两个迈克，但现在在这两个迈克中，我想以降序的形式对他们的薪水进行排序，这样新的结果将是:

<name: Andy salary: 60000 year of birth: 1966>, 
    <name: Gemasio salary: 60000 year of birth: 1971>, 
    <name: Joe salary: 100000 year of birth: 1980>, 
    <name: Mike salary: 80000 year of birth: 1970>, 
    <name: Mike salary: 90000 year of birth: 1992>

除了迈克，我不想更改其他订单

但是我无法得到任何人想要的结果 请帮助我找出问题所在或我需要做些什么才能从这里走得更远。谢谢:)

Answer 1

你可能正在寻找

.sorted(Comparator
        .comparing(Employee::getName)         //sort by name
        .thenComparing(Employee::getSalary))  //in case of same names sort by salary

如果你想使用降序，你可以创建单独的 Comparator，它将根据 Employee::getSalary 进行比较，并使用它的 reversed() 版本如下:

        .sorted(Comparator
                .comparing(Employee::getName)
                .thenComparing(
                        Comparator.comparing(Employee::getSalary).reversed()
                )
        )

groupingBy(Employee::getName) 可能存在的问题是它返回无序的 HashMap。如果你想确保键的顺序基于它的插入(你可能是，否则之前根据名称排序是没有意义的)你可以使用:

.collect(Collectors.groupingBy(Employee::getName, LinkedHashMap::new, Collectors.toList()));

演示:

//code creating list
//...

//our code
Map<String, List<Employee>> grouped = list.stream()
        .filter(e -> e.getSalary() > 55000)
        .filter(e -> e.getSalary() < 120000)
        .sorted(Comparator
                .comparing(Employee::getName)
                .thenComparing(
                        Comparator.comparing(Employee::getSalary).reversed()
                )
        )
        .collect(Collectors.groupingBy(Employee::getName, LinkedHashMap::new, Collectors.toList()));

for (Map.Entry<String, List<Employee>> entry : grouped.entrySet()) {
    System.out.println(entry.getKey());
    for (Employee emp : entry.getValue()) {
        System.out.println("\t" + emp);
    }
}

输出:

Andy
    Employee [name=Andy, salary=60000, yearOfBirth=1966]
Gemasio
    Employee [name=Gemasio, salary=60000, yearOfBirth=1971]
Joe
    Employee [name=Joe, salary=100000, yearOfBirth=1980]
Mike
    Employee [name=Mike, salary=90000, yearOfBirth=1970]
    Employee [name=Mike, salary=80000, yearOfBirth=1992]