Map Triples to IDs/numbers using ZipWithIndex/ZipWithUniqueID - scala

Question

I asked the question before but it was unclear so I added more explanation to be more clear and to get help.
replace strings with ZipWithIndex/ZipWithUniqueID
I am trying to map string to number using ZipWithIndex OR ZipWithUniqueID
lets say I have this format
("u1",("name", "John Sam"))
("u2",("age", "twinty Four"))
("u3",("name", "sam Blake"))
I want this result
(0,(3,4))
(1,(5,6))
(2,(3,8))
I tried to use zipWithIndex directly to the triples but I got each letter mapped to a number I want to map the whole string without dividing it.
and tried to extract the first element in the key, value pair
so I did
val first = file.map(line=> line._1).distinct()
then apply ZipWithIndex
val z1= first.ZipWithIndex()
I got result like this
("u1",0)
("u2",1)
("u3",2)
now I need to take the ids/numbers and change it in my original file. and I need to keep all the distinct ids/numbers in hashTable to be able to look for them later on.
is there any way to do that? Any suggestions?
I hope you got my question

Answer 1

You mean something like this?
val file = List(("u1",("name", "John Sam")),
("u2",("age", "twinty Four")),
("u3",("name", "sam Blake")))
val first = file.map(line=> line._1) ++
file.flatMap(line=> List(line._2._1, line._2._2)).distinct
val z1: Map[String,Int] = Map[String,Int](first.zipWithIndex:_*)
file.map{ l =>
(z1(l._1),
(z1(l._2._1), z1(l._2._2)))
}