Question

我有一个名为Student的类，有三个字段，我需要将该类转换为GroupedStudent。如何实现这一目标？

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我需要将上面的列表转换为GroupedStudent List，我该怎么做？

它们应按studentName分组，并将SubjectMarks列为列表。

var students: List<Student> = mutableListOf(Student("A", "X", 1), Student("A", "Y", 2), Student("B", "X", 2), Student("B", "Y", 2))

Answer 1

使用Kotlin stdlib Collections functions：

var students: List<Student> = mutableListOf(Student("A", "X", 1), Student("A", "Y", 2), Student("B", "X", 2), Student("B", "Y", 2))

val groupedMap = students.groupBy { it.name }
val groupedStudents = mutableListOf<GroupedStudent>()
groupedMap.forEach { key, value ->
    val groupedStudent = GroupedStudent(key, value.map { SubjectMarks(it.subject, it.marks) }.toMutableList())
    groupedStudents.add(groupedStudent)
}

这将产生List GroupedStudent，每个GroupedStudent包含一个学生列表＆＃39}。标记谁有这个名字。

Answer 2

如果您的Student类是正确的，即标记应该是Int而不是数组或整数列表，并且您真的想将其转换为＆＃34; GroupedStudent List＆＃34;你可以使用sortedWith和map方法：

class Student(var name: String, var subject: String, var marks: Int) {
  override fun toString(): String = "name: $name, subject: $subject, marks: $marks"
}

class GroupedStudent(var name: String, var subMarks: MutableList<SubjectMarks>) {
  override fun toString(): String = "name: $name, subject: $subMarks"
}

class SubjectMarks(var subject: String, var marks: Int) {
  override fun toString(): String = "subject name: $subject, marks: $marks"
}

fun main(args: Array<String>) {
  val students: List<Student> = mutableListOf(Student("B", "Y", 2),
                                              Student("A", "X", 1),
                                              Student("B", "X", 2),
                                              Student("A", "Y", 2))

  val groupedStudents: List<GroupedStudent> = students.sortedWith(compareBy({ it.name }))
        .map { student -> GroupedStudent(student.name, mutableListOf(SubjectMarks(student.subject, student.marks))) }

  println(groupedStudents)
}

result：

[name: A, subject: [subject name: X, marks: 1], name: A, subject: [subject name: Y, marks: 2], name: B, subject: [subject name: Y, marks: 2], name: B, subject: [subject name: X, marks: 2]]

但是如果要将学生对象分组到名称为关键的地图中，并将主题/标记列表作为对子列表，则可以使用groupBy：

class Student(var name: String, var subject: String, var marks: Int) {
  override fun toString(): String = "name: $name, subject: $subject, marks: $marks"
}

class GroupedStudent(var name: String, var subMarks: MutableList<SubjectMarks>) {
  override fun toString(): String = "name: $name, subject: $subMarks"
}

class SubjectMarks(var subject: String, var marks: Int) {
  override fun toString(): String = "subject name: $subject, marks: $marks"
}

fun main(args: Array<String>) {
  val students: List<Student> = mutableListOf(Student("B", "Y", 2),
                                          Student("A", "X", 1),
                                          Student("B", "X", 2),
                                          Student("A", "Y", 2))

  val groupedStudentsMap = mutableMapOf<String, List<Pair<String, Int>>>()
  students.groupBy { it.name }
        .forEach({ (key, value) -> 
          groupedStudentsMap[key] = value.map { element -> Pair(element.subject, element.marks) } })

  println(groupedStudentsMap)
}

result：

{B=[(Y, 2), (X, 2)], A=[(X, 1), (Y, 2)]}

Answer 3

您可以按学生的姓名分组，然后执行两个简单的映射：

students.groupBy(Student::name).map { (name, students) ->
    GroupedStudent(name, students.map { SubjectMarks(it.subject, it.marks) }.toMutableList())
}

在kotlin中分组课程

3 个答案: