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一些列表,它执行一些过滤,例如:salary> x和salary <y并根据员工姓名排序,然后再次排序,如果多个员工有相同的名字但工资不同但现在使用他们的工资

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:排序后只有我得到的名字::

<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我想以分层形式对他们的工资进行排序,以便新的结果将是:

<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>

除了迈克我不想改变其他订单

但我无法得到理想的结果任何人都可以帮助我解决出错的问题或我需要做些什么来进一步从这里开始.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comparing基于Employee::getSalary并使用其reversed()版本,如:

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

可能的问题groupingBy(Employee::getName)是它返回无序的HashMap.如果你想确保基于其插入的键的顺序(你可能是,否则根据名称排序没有任何意义)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]