CODEWITHSUNDEEP
scala
Teja
Teja

Reputation: 466

How to map id and name to the given sequence in scala

How to map id and name to the given sequence in scala ... If suppose I have Input Sequence like this 

Seq((1, "teja"), (2, "mahesh"))

and I need to get Output like 

Map[("id" ->1 ,"name"->"teja"),("id" ->2,"name"->"mahesh")]

How to do it?

Upvotes: 1

Views: 711

Answers (1)

Andreas Neumann
Andreas Neumann

Reputation: 10884

You can do it like this:

scala> Seq((1, "teja"), (2, "mahesh")) map { case (id,name) => ("id" -> id, "name" -> name)  }
res0: Seq[((String, Int), (String, String))] = List(((id,1),(name,teja)), ((id,2),(name,mahesh)))

to keep the -> in the toString use a Map

scala> Seq((1, "teja"), (2, "mahesh")) map { case (id,name) => Map("id" -> id, "name" -> name)  }
res1: Seq[scala.collection.immutable.Map[String,Any]] = List(Map(id -> 1, name -> teja), Map(id -> 2, name -> mahesh))

if you need it within a map you can do 

scala> Seq((1, "teja"), (2, "mahesh")).map { case (id,name) => ("id" -> id, "name" -> name)  }.toMap
res2: scala.collection.immutable.Map[(String, Int),(String, String)] = Map((id,1) -> (name,teja), (id,2) -> (name,mahesh))

then the key will be a Tuple2[String,Int] though.

Upvotes: 2

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