How to sort contents of case class - scala

Question

If I have a case class like below:
case class Student(name: String, activities: Seq[String], grade: String)
And I have a List like this:
val students = List(
Student("John", List("soccer", "Video Games"), "9th"),
Student("Jane", List("sword fighting", "debate"), "10th"),
Student("Boy Wonder", List("1", "5", "2"), "5th")
)
How can I sort the contents based on name and activities attributes to form a string? In the scenario above the string would be:
boywonder_1_2_5_5th_jane_debate_swordfighting_10th_john_soccer_videogames_9th
The sorting in this case is done like this:
First the elements are sorted with name -- Thats why in the final string boywonder comes first
Then that elements' activities are sorted as well -- Thats why Boy Wonder's activities are sorted as 1_2_5

Answer 1

You need to:
Make everything lowercase.
Sort the inner list activities.
Sort the outer list students, by name.
Turn everything into a String.
Here is the code.
students
.map { student =>
student.copy(
name = student.name.toLowerCase,
activities = student.activities.sorted.map(activity => activity.toLowerCase)
)
}.sortBy(student => student.name)
.map(student => s"${student.name}${student.activities.mkString}${student.grade}")
.mkString
.replaceAll("\\s", "")
// res: String = "boywonder1255thjanedebateswordfighting10thjohnvideogamessoccer9th"