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.
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3Do you want b to map to 2 because it's the second in the Set? Or becasue it's the second letter in the alphabet? Sets aren't ordered, as other posters have pointed out, so the first won't work. – The Archetypal Paul Jan 31 '11 at 14:54
4 Answers
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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.
Adam Rabung
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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))
Frank S. Thomas
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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
Xavier Guihot
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GClaramunt
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