.collect(Collectors.toList()) 和 Streams on Java 方法
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.collect(Collectors.toList()) and Streams on Java Method
提问by Giuseppe Canto
I have a collection (as hashmap) of Doctors, into a generic Hospital class.
我有一个医生的集合（作为哈希图），放到一个通用的医院类中。
Map<Integer, Doctor> doctors = new HashMap<Integer, Doctor>();
For each doctor I have some information such as in the class code (focus on the patients):
对于每个医生，我都有一些信息，例如在类代码中（专注于患者）：
public class Doctor extends Person {
private int id;
private String specialization;
private List<Person> patients = new LinkedList<Person>();
My purpose is to write this function which return busy doctors: doctors that has a number of patients larger than the average.
我的目的是编写这个函数来返回忙碌的医生：患者数量大于平均水平的医生。
/**
* returns the collection of doctors that has a number of patients larger than the average.
*/
Collection<Doctor> busyDoctors(){
Collection<Doctor> doctorsWithManyPatients =
doctors.values().stream()
.map( doctor -> doctor.getPatients() )
.filter( patientsList -> { return patientsList.size() >= AvgPatientsPerDoctor; })
.collect(Collectors.toList());
return null;
}
I want to use the streams as above to perform this operation. The problem is in collectmethod because at that point of usage doctorsWithManyPatientsis of type List<Collection<Person>>and not Collection<Doctor>. How could I do that?
我想使用上述流来执行此操作。问题出在collect方法上，因为在那个时候使用的doctorsWithManyPatients是 typeList<Collection<Person>>而不是Collection<Doctor>。我怎么能那样做？
Assume that AvgPatientsPerDoctoris already defined somewhere.
假设AvgPatientsPerDoctor已经在某处定义了。
回答by Andrew Tobilko
You needn't use map(Doctor -> List<Person>), it will be used in the filter:
您不需要使用map( Doctor -> List<Person>)，它将用于filter：
doctors
.values()
.stream()
.filter( d -> d.getPatients().size() >= AvgPatientsPerDoctor)
.collect(Collectors.toList());
For your case, map( doctor -> doctor.getPatients() )returns Stream<List<Person>>and you should convert it to Stream<Doctor>again after filtering and before calling the collectmethod.
对于您的情况，map( doctor -> doctor.getPatients() )返回Stream<List<Person>>并且您应该Stream<Doctor>在filtering之后和调用该collect方法之前再次将其转换为。
There is a different way that isn't the best one. Keep in mind that it changes the origin collection.
有一种不同的方式并不是最好的方式。请记住，它会更改原始集合。
doctors.values().removeIf(d -> d.getPatients().size() < AvgPatientsPerDoctor);
