CODEWITHSUNDEEP
scalascala-collections
Torsten Schmidt
Torsten Schmidt

Reputation: 217

Scala Convert Set to Map

How do I convert a Set("a","b","c") to a Map("a"->1,"b"->2,"c"->3)? I think it should work with toMap.

Upvotes: 21

Views: 21425

Answers (4)

GClaramunt
GClaramunt

Reputation: 3158

toMap only works if the Set entries are key/value pairs (e.g. Set(("a",1),("b",2),("c",3))).

To get what you want, use zipWithIndex:

Set("a","b","c") zipWithIndex
// Set[(String, Int)] = Set((a,0), (b,1), (c,2))

or (as in you original question):

Set("a","b","c") zip (1 to 3) toMap

Upvotes: 7

Viktor Nordling
Viktor Nordling

Reputation: 9284

This would also work:

(('a' to 'c') zip (1 to 3)).toMap

Upvotes: 3

Frank S. Thomas
Frank S. Thomas

Reputation: 4775

Here is another solution that uses a Stream of all natural numbers beginning from 1 to be zipped with your Set:

scala> Set("a", "b", "c") zip Stream.from(1) toMap
Map((a,1), (b,2), (c,3))

Upvotes: 19

Adam Rabung
Adam Rabung

Reputation: 5224

zipWithIndex is probably what you are looking for. It will take your collection of letters and make a new collection of Tuples, matching value with position in the collection. You have an extra requirement though - it looks like your positions start with 1, rather than 0, so you'll need to transform those Tuples:

Set("a","b","c")
  .zipWithIndex    //(a,0), (b,1), (c,2)
  .map{case(v,i) => (v, i+1)}  //increment each of those indexes
  .toMap //toMap does work for a collection of Tuples

One extra consideration - Sets don't preserve position. Consider using a structure like List if you want the above position to consistently work.

Upvotes: 31

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