How to convert list of list to simple list by removing duplicate values using scala?

Question

I have following list -

List(List(
List(((groupName,group1),(tagMember,["192.168.20.30","192.168.20.20","192.168.20.21"]))), 
List(((groupName,group1),(tagMember,["192.168.20.30"]))),
List(((groupName,group1),(tagMember,["192.168.20.30","192.168.20.20"])))))

I want to convert it to -

List((groupName, group1),(tagMember,["192.168.20.30","192.168.20.20","192.168.20.21"]))

I tried to use .flatten but unable to form desired output.

How do I get above mentioned output using scala??

Can you simplify your example into some code we can copy into a repl? And maybe use shorter strings? — Chris Martin
– Chris Martin, Commented Oct 20, 2014 at 6:26
I used list.flatten to remove flatten list but have no idea how to merge list by removing duplicate elements — Vishwas
– Vishwas, Commented Oct 20, 2014 at 6:32
Is the IP address the only variable part in the list, or could there be other entries in the list like (groupName, group2)? — Deepti Bhatia
– Deepti Bhatia, Commented Oct 20, 2014 at 8:37

Deepti Bhatia · Accepted Answer · 2014-10-20 09:29:30Z

I had to make some changes to your input to make it valid.

Input List:

   val ll = List(List(
      List((("groupName","group1"),("tagMember", List("192.168.20.30","192.168.20.20","192.168.20.21")))),
      List((("groupName","group1"),("tagMember",List("192.168.20.30")))),
      List((("groupName","group1"),("tagMember",List("192.168.20.30","192.168.20.20"))))
    ))

Code below works if the group, and tagMember are the same across all the elements in the list

  def getUniqueIpsConstantGroupTagMember(inputList:  List[List[List[((String, String), (String, List[String]))]]]) = {
      // List[((String, String), (String, List[String]))]
      val flattenedList  = ll.flatten.flatten

      if (flattenedList.size > 0) {
         val group = flattenedList(0)._1
         val tagMember = flattenedList(0)._2._1
         val ips = flattenedList flatMap (_._2._2)

         ((group), (tagMember, ips.distinct))
      } 
         else List()
  }

  println(getUniqueIpsConstantGroupTagMember(ll))

Output:

((groupName,group1),(tagMember,List(192.168.20.30, 192.168.20.20, 192.168.20.21)))

Now, let's assume you could have different groupNames.

Sample input:

  val listWithVariableGroups = List(List(
      List((("groupName","group1"),("tagMember",List("192.168.20.30","192.168.20.20","192.168.20.21")))),
      List((("groupName","group1"),("tagMember",List("192.168.20.30")))),
      List((("groupName","group1"),("tagMember",List("192.168.20.30","192.168.20.20")))),
      List((("groupName","group2"),("tagMember",List("192.168.20.30","192.168.20.10"))))   
    ))

The following code should work.

  def getUniqueIpsForMultipleGroups(inputList:  List[List[List[((String, String), (String, List[String]))]]]) = {
    val flattenedList = inputList.flatten.flatten

    // Map[(String, String),List[(String, List[String])]]
    val groupedByGroupNameId = flattenedList.groupBy(p => p._1) map {
      case (key, value) => (key, ("tagMember", extractUniqueTagIps(value)))
    }
    groupedByGroupNameId
  }

  def extractUniqueTagIps(list: List[((String, String), (String, List[String]))]) = {
    val ips = list flatMap (_._2._2)
    ips.distinct
  }

  getUniqueIpsForMultipleGroups(listWithVariableGroups).foreach(println)

Output:

((groupName,group1),(tagMember,List(192.168.20.30, 192.168.20.20, 192.168.20.21)))
((groupName,group2),(tagMember,List(192.168.20.30, 192.168.20.10)))

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