With regard to maps in Scala, if ms - (k, 1, m)
returns the map containing all mappings of ms except for any mapping with the given keys, x, 1 and m.
Then, what statement will return a map of all mappings of ms with only the given keys, x, 1 and m. i.e. I'm looking for the subset of ms where only k, 1 and m are keys.
This works, but it is terrible:
scala> val originalMap = Map("age" -> "20", "name" -> "jack", "hobby" -> "jumping")
ms: scala.collection.immutable.Map[java.lang.String,java.lang.String] = Map(age -> 20, name -> jack, hobby -> jumping)
scala> val interestingKeys = List("name", "hobby")
interesting: List[java.lang.String] = List(name, hobby)
scala> val notInterestingMap = originalMap -- interestingKeys
notInterestingMap: scala.collection.immutable.Map[java.lang.String,java.lang.String] = Map(age -> 20)
scala> val interestingMap = originalMap -- notInterestingMap.keySet
interestingMap: scala.collection.immutable.Map[java.lang.String,java.lang.String] = Map(name -> jack, hobby -> jumping)
filterKeys
can help:
scala> originalMap.filterKeys(interestingKeys.contains)
res0: scala.collection.immutable.Map[java.lang.String,java.lang.String] = Map(name -> jack, hobby -> jumping)
Because filterKeys
filters based on an arbitrary predicate, it has to consider every key in the map. This might be fine or not, depending on how large the map is, etc., but it's definitely not necessary for the operation you describe. I'd use something like the following:
interestingKeys.flatMap(k => originalMap.get(k).map((k, _))).toMap
This will be O(n)
or O(n log m)
depending on your map implementation (where n
is the size of interestingKeys
and m
is the size of the map), instead of O(m log n)
or O(mn)
.
If you really want your ~
operator, you can use the pimp-my-library pattern:
class RichMap[A, B](m: Map[A, B]) {
def ~(ks: A*) = ks.flatMap(k => m.get(k).map((k, _))).toMap
}
implicit def enrichMap[A, B](m: Map[A, B]) = new RichMap(m)
Now originalMap ~ ("name", "hobby")
returns Map(name -> jack, hobby -> jumping)
, as you'd expect.
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