How to transform two Maps into a single Map

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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"))
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

Merging Two Maps with Java 8, First, we transform map1 and map2 into a single stream. Next, we convert the stream into the map. As we can see, the last argument of toMap() is� First, we transform map1 and map2 into a single stream. Next, we convert the stream into the map. As we can see, the last argument of toMap () is a merging function. It solves the duplicate keys problem by picking the id field from v1 entry, and the name from v2.

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

Merge Lists of Map to Map, Java 8 Style!, To calculate the result, we will first convert the list of maps to stream of In reduce we will get two maps to be merged, where we will use Convert both the maps to streams, and concat the stream to get a single stream of� To do this, double-click the map's name in the Contents pane, and rename the map in the Map Properties dialog box. Repeat for the second map. On the Insert tab, click the New Layout drop-down menu, and select the desired layout for the project. The Layout is added to the data frame.

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)

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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)

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Or idemponent for original maps solution with third map: Map<String, Integer> map3 = new HashMap<String, Integer>(map2); map1.forEach(map3::putIfAbsent); And here is a way to merge two maps into fast immutable one with Guava that does least possible intermediate copy operations:

You can transform an existing map from one coordinate system to another by querying the objects from the attached source drawing into the current drawing. To transform the coordinate system of a drawing Open a new drawing. Attach the drawing whose coordinate system you want to transform.

Comments
  • What's supposed to happen if they are different lengths? Ignore the key and/or value if it can't be paired?
  • In those cases, the pair should be ignored.
  • You can just do v2 <- m1.get(k), delete the if and use v2 rather than v2.get
  • that only works if the order of the map is the same!! Check scalafiddle.io/sf/5F2Hc3P/0
  • this only fails now for the last comment of d-_-b: "In those cases, the pair should be ignored." - check scalafiddle.io/sf/5F2Hc3P/4
  • this only works if you both Maps have exactly the same keys - check scalafiddle.io/sf/5F2Hc3P/1