Follow up to this question.
How to reorder
List(Map(d -> 4, a -> 12, c -> 30, b -> 2), Map(d -> 80, a -> 22, c -> 6666, b -> 442))
to
List(Map(d -> 4, b -> 2, a -> 12, c -> 30), Map(d -> 80, b -> 442, a -> 22, c -> 6666))
using Scala?
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2 Answers
A Map is usually an unordered collection. To be more precise, a Map is a Set of keys with values attached to them. Since Sets only have distinct elements, and aren't ordered, so are Maps.
SortedMap[String, Int]
However, there is an implementation of Map with sorted keys, and that's SortedMap. It uses an implicit Ordering[Key] to keep key-value pairs ordered by their key (note that it keeps unicity of keys, so that you cannot have two values for a given key).
For instance
SortedMap(-1 -> "neg", 1 -> "pos") + (0 -> "zero") ==
SortedMap(-1 -> "neg", 0 -> "zero", 1 -> "pos")
However, in your case, it might be difficult to keep an implicit Ordering[String] different from the usual one (which is lexicographical order).
Seq[(String, Int)]
You might want to convert your Map[String, Int] to a Seq[(String, Int)]. That way, you can play with the order of your values, but you'll loose the fast lookup for a given key.
If you want to do this, you can specify your order first by using a list of keys in order, and then mapping on it using the original map.
val original: Map[String, Int] = Map("d" -> 4, "a" -> 12, "c" -> 30, "b" -> 2)
orderedMap: List[(String, Int)] = for {
key <- List("d", "b", "a", "c")
value <- original.get(key)
} yield key -> value
Note that I used original.get(key) to avoid the case where one of your keys is not actually defined in the original map.
Now, to get a value for a given key, using this List[(String, Int)] type, you have to do
def getKey(key: String): Option[Int] = orderedMap.
find(_._1 == key). //find the first pair where the first element is the desired key
map(_._2) // keep only the value if it exists.
Comments
SortedMap.apply take implicit Ordering that define order. For example:
implicit val stringReverse = Ordering.String.reverse
val xs = Map("a" -> 3, "b" -> 5)
SortedMap(xs.toList:_*) // Map(b -> 5, a -> 3)
