How to combine all the values with the same key in Scala? - scala

Question

I have a map like :
val programming = Map(("functional", 1) -> "scala", ("functional", 2) -> "perl", ("orientedObject", 1) -> "java", ("orientedObject", 2) -> "C++")
with the same first element of key appearing multiple times.
How to regroup all the values corresponding to the same first element of key ? Which would turn this map into :
Map("functional" -> List("scala","perl"), "orientedObject" -> List("java","C++"))

Answer 1

UPDATE: This answer is based upon your original question. If you need the more complex Map definition, using a tuple as the key, then the other answers will address your requirements. You may still find this approach simpler.
As has been pointed out, you can't actually have multiple keys with the same value in a map. In the REPL, you'll note that your declaration becomes:
scala> val programming = Map("functional" -> "scala", "functional" -> "perl", "orientedObject" -> "java", "orientedObject" -> "C++")
programming: scala.collection.immutable.Map[String,String] = Map(functional -> perl, orientedObject -> C++)
So you end up missing some values. If you make this a List instead, you can get what you want as follows:
scala> val programming = List("functional" -> "scala", "functional" -> "perl", "orientedObject" -> "java", "orientedObject" -> "C++")
programming: List[(String, String)] = List((functional,scala), (functional,perl), (orientedObject,java), (orientedObject,C++))
scala> programming.groupBy(_._1).map(p => p._1 -> p._2.map(_._2)).toMap
res0: scala.collection.immutable.Map[String,List[String]] = Map(functional -> List(scala, perl), orientedObject -> List(java, C++))

Answer 2

Based on your edit, you have a data structure that looks something like this
val programming = Map(("functional", 1) -> "scala", ("functional", 2) -> "perl",
("orientedObject", 1) -> "java", ("orientedObject", 2) -> "C++")
and you want to scrap the numerical indices and group by the string key. Fortunately, Scala provides a built-in that gets you close.
programming groupBy { case ((k, _), _) => k }
This will return a new map which contains submaps of the original, grouped by the key that we return from the "partial" function. But we want a map of lists, so let's ignore the keys in the submaps.
programming groupBy { case ((k, _), _) => k } mapValues { _.values }
This gets us a map of... some kind of Iterable. But we really want lists, so let's take the final step and convert to a list.
programming groupBy { case ((k, _), _) => k } mapValues { _.values.toList }

Answer 3

You should try the .groupBy method
programming.groupBy(_._1._1)
and you will get
scala> programming.groupBy(_._1._1)
res1: scala.collection.immutable.Map[String,scala.collection.immutable.Map[(String, Int),String]] = Map(functional -> Map((functional,1) -> scala, (functional,2) -> perl), orientedObject -> Map((orientedObject,1) -> java, (orientedObject,2) -> C++))
you can now "clean" by doing something like:
scala> res1.mapValues(m => m.values.toList)
res3: scala.collection.immutable.Map[String,List[String]] = Map(functional -> List(scala, perl), orientedObject -> List(java, C++))

Answer 4

Read the csv file and create a map that contains key and list of values.
val fileStream = getClass.getResourceAsStream("/keyvaluepair.csv")
val lines = Source.fromInputStream(fileStream).getLines
var mp = Seq[List[(String, String)]]();
var codeMap=List[(String, String)]();
var res = Map[String,List[String]]();
for(line <- lines )
{
val cols=line.split(",").map(_.trim())
codeMap ++= Map(cols(0)->cols(1))
}
res = codeMap.groupBy(_._1).map(p => p._1 -> p._2.map(_._2)).toMap

Answer 5

Since no one has put in the specific ordering he asked for:
programming.groupBy(_._1._1)
.mapValues(_.toSeq.map { case ((t, i), l) => (i, l) }.sortBy(_._1).map(_._2))