How to transform two Maps into a single Map - scala

Question

Suppose I have two Maps like these
Map(("a" -> "x-100"), ("b" -> "x-200"), ("c" -> "x-300"))
Map(("a" -> "y-100"), ("b" -> "y-200"), ("c" -> "y-300"))
What would be the simplest way to transform them into the following (assuming that all the values are unique and with possibly different lengths)?
Map(("x-100" -> "y-100"), ("x-200" -> "y-200"), ("x-300" -> "y-300"))

Answer 1

Or with a for comprehension and some safety:
for{
(k, v1) <- m1
v2 = m2.get(k)
if (v2.isDefined)
} yield (v1 -> v2.get)
This returns Map(x-100 -> y-100, x-200 -> y-200, x-300 -> y-300)
Check ScalaFiddle

Answer 2

val m1 = Map(("a" -> "x-100"), ("b" -> "x-200"), ("c" -> "x-300"),("d" -> "ignored"))
val m2 = Map(("a" -> "y-100"), ("b" -> "y-200"), ("c" -> "y-300"))
m1.keySet.intersect(m2.keySet).map(k=>m1(k)->m2(k)).toMap

Answer 3

You can do it like this:
val a = Map(("a" -> "x-100"), ("b" -> "x-200"), ("c" -> "x-300"))
val b = Map(("a" -> "y-100"), ("b" -> "y-200"), ("c" -> "y-300"))
def zipMaps(map1: Map[String, String], map2: Map[String, String]) = {
for(key <- map1.keys ++ map2.keys)
yield (key, map1.get(key), map2.get(key))
}
val result = zipMaps(a,b).map{
case (k,Some(v1),Some(v2)) => (v1,v2)
case (k,_ ,_)=> ("", "")
}.toMap.filterKeys(_ != "")
// result = Map(x-100 -> y-100, x-200 -> y-200, x-300 -> y-300)

Answer 4

val a = Map("a" -> "x-100", "b" -> "x-200", "c" -> "x-300")
val b = Map("a" -> "y-100", "b" -> "y-200", "c" -> "y-300")
val c = a.map {
case (k, v) => v -> b(k)
}
println(c) // Map(x-100 -> y-100, x-200 -> y-200, x-300 -> y-300)