如何以正确的方式过滤 ArrayList 的内容？答案

【问题标题】：How to filter the content of an ArrayList the right way?如何以正确的方式过滤 ArrayList 的内容？
【发布时间】：2019-09-08 23:19:17
【问题描述】：

我正在努力寻找一种更好的方法来过滤 ArrayList 的内容

例如，这个程序有一个名为“students”的主 ArrayList，然后我从这个列表的内容中创建了其他子列表（oldStudents、youngStudents、stupidStudents、smartStudents。目标是根据学生的用户选择过滤 ArrayList （年轻、年老、聪明或愚蠢）

    ArrayList<String> students = new ArrayList<String>(); 
    ArrayList<String> smartStudents = new ArrayList<String>(); 
    ArrayList<String> stupidStudents = new ArrayList<String>(); 
    ArrayList<String> oldStudents = new ArrayList<String>(); 
    ArrayList<String> youngStudents = new ArrayList<String>(); 


    //adding all the students to students list 
    Collections.addAll(students, "Ram", "Mohan", "Sohan", "Rabi", "Shabbir","Jack", "Johnson", "Peter", "Despina", "Me");
    //adding young students to youngStudents list 
    Collections.addAll(youngStudents, "Ram", "Mohan", "Sohan", "Rabi", "Shabbir");
    //adding smart students to oldStudents list 
    Collections.addAll(oldStudents, "Jack", "Johnson", "Peter", "Despina", "Me");
    //adding smart students to smartStudents list 
    Collections.addAll(smartStudents, "Sohan", "Rabi", "Peter", "Despina");
    //adding smart students to stupidStudents list 
    Collections.addAll(stupidStudents, "Ram", "Mohan", "Shabbir","Jack", "Johnson", "Me");

    Scanner input = new Scanner(System.in);
    String uInput = "";



    System.out.print("This is a students search engine, write 'young' for younger students and 'old' for older ones ");
    uInput = input.nextLine();

    if(uInput.equals("young")) {

        students.removeAll(oldStudents); 

    } else if (uInput.equals("old")) {

        students.removeAll(youngStudents);

    }


    System.out.print("now write 'Smart' for smarter students and 'Stupid' for less smart students ");
    uInput = input.nextLine();

  if(uInput.equals("smart")) {

        students.removeAll(stupidStudents); 

    } else if (uInput.equals("Stupid")) {

        students.removeAll(smartStudents);

    }


  System.out.println(students);

它正在工作，但我相信有更好的方法来实现这一点

【问题讨论】：

第一次将所有这些列表插入学生列表中...
除非您需要在类中包含其他功能，否则最好分配给接口类型。您还可以通过执行List<String> stupid = new ArrayList<>(List.of("stupid1", "stupid2")); 之类的操作来创建可变列表。还有其他使用streams() 和收集器的方法。

标签： java search arraylist filter

【解决方案1】：

您可以使用这种方法：

class Student { String name; boolean old; boolean stupid; }

List<Student> students = initializeListOfStudents();

// alter these flags accordingly to your needs
Boolean old = null;
Boolean stupid = true;

// then do the filter
List<Student> allStupids = students.stream()
    .filter(it -> (old==null || it.old == old) && (stupid==null || it.stupid==stupid))
    .collect(Collectors.toList());

【讨论】：

【解决方案2】：

Java 是一种面向对象的语言。使用它。

创建一个包含 3 个字段的 Student 类：String name、boolean old 和 boolean smart。添加 getter 方法。

您现在可以轻松过滤，例如

// New list with dumb students
dumpStudents = studentList.stream().filter(s -> ! s.isSmart()).collect(Collectors.toList());

// Remove old people from list
studentList.removeIf(Student::isOld);

【讨论】：

【解决方案3】：

在 Java 8 中，您可以使用 Stream 来过滤列表。在此链接中，您可以找到不同的示例：

https://www.baeldung.com/java-stream-filter-lambda

【讨论】：