I am new to scala. I have a Map. I want to set a value in the Map with a particular key. Here is the code I am writing -
var mp: Map[Int, ParticipationStateTransition] = Map.empty[Int, ParticipationStateTransition]
val change: ParticipationStateTransition = new ParticipationStateTransition
mp(ri.userID) = change
The error it is showing me on the third line is -
application does not take parameters
What am I doing wrong? Thanks in advance.
Use .updated :
scala> val m = Map(1 -> 2)
m: scala.collection.immutable.Map[Int,Int] = Map(1 -> 2)
scala> val n = m.updated(1, 3)
n: scala.collection.immutable.Map[Int,Int] = Map(1 -> 3)
scala> m
res0: scala.collection.immutable.Map[Int,Int] = Map(1 -> 2)
scala> n
res1: scala.collection.immutable.Map[Int,Int] = Map(1 -> 3)
Note that scala's Map are immutable, so you need to assign the return value of .updated, it will not change the original map.
If you want to change the map in place, you can use collection.mutable.Map and then
scala> val m = collection.mutable.Map(1 -> 2)
m: scala.collection.mutable.Map[Int,Int] = Map(1 -> 2)
scala> m.update(1, 3)
scala> m
res3: scala.collection.mutable.Map[Int,Int] = Map(1 -> 3)
If you want to set multiple values at once, you can do :
scala> val m = Map(1 -> 2)
m: scala.collection.immutable.Map[Int,Int] = Map(1 -> 2)
scala> val n = m ++ List((1 -> 3), (2 -> 4)) // also accepts an Array, a Map, …
n: scala.collection.immutable.Map[Int,Int] = Map(1 -> 3, 2 -> 4)
Related
The example in the README is very elegant:
scala> Map(1 -> Max(2)) + Map(1 -> Max(3)) + Map(2 -> Max(4))
res0: Map[Int,Max[Int]] = Map(2 -> Max(4), 1 -> Max(3))
Essentially the use of Map here is equivalent to SQL's group by.
But how do I do the same with an arbitrary Aggregator? For example, to achieve the same thing as the code above (but without the Max wrapper class):
scala> import com.twitter.algebird._
scala> val mx = Aggregator.max[Int]
mx: Aggregator[Int,Int,Int] = MaxAggregator(scala.math.Ordering$Int$#78c77)
scala> val mxOfMap = // what goes here?
mxOfMap: Aggregator[Map[Int,Int],Map[Int,Int],Map[Int,Int]] = ...
scala> mxOfMap.reduce(List(Map(1 -> 2), Map(1 -> 3), Map(2 -> 4)))
res0: Map[Int,Int] = Map(2 -> 4, 1 -> 3)
In other words, how to I convert (or "lift") an Aggregator that operates on values of type T into an Aggregator that operates on values of type Map[K,T] (for some arbitrary K)?
Looks like this can be done fairly easily for Semigroup at least. This should be sufficient in the case where there is no additional logic in the "compose" or "present" phases of the aggregator which needs to be preserved (a Semigroup can be obtained from an Aggregator, discarding compose/prepare).
The code to answer the original question is:
scala> val sgOfMap = Semigroup.mapSemigroup[Int,Int](mx.semigroup)
scala> val mxOfMap = Aggregator.fromSemigroup(sgOfMap)
scala> mxOfMap.reduce(List(Map(1 -> 2), Map(1 -> 3), Map(2 -> 4)))
res0: Map[Int,Int] = Map(2 -> 4, 1 -> 3)
But in practice, it would be better to start by constructing the arbitrary Semigroup directly, rather than constructing an Aggregator merely to extract the semigroup from:
scala> import com.twitter.algebird._
scala> val mx = Semigroup.from { (x: Int, y: Int) => Math.max(x, y) }
scala> val mxOfMap = Semigroup.mapSemigroup[Int,Int](mx)
scala> mxOfMap.sumOption(List(Map(1 -> 2), Map(1 -> 3), Map(2 -> 4)))
res33: Option[Map[Int,Int]] = Some(Map(2 -> 4, 1 -> 3))
Alternatively, convert to aggregator: Aggregator.fromSemigroup(mxOfMap)
I have a list:
List(1,2,3,4,5,6)
that I would like to to convert to the following map:
Map(1->2,3->4,5->6)
How can this be done?
Mostly resembles #Vakh answer, but with a nicer syntax:
val l = List(1,2,3,4,5,6)
val m = l.grouped(2).map { case List(key, value) => key -> value}.toMap
// Map(1 -> 2, 3 -> 4, 5 -> 6)
Try:
val l = List(1,2,3,4,5,6)
val m = l.grouped(2).map(l => (l(0), l(1))).toMap
if the list is guaranteed to be of even length:
val l = List(1,2,3,4,5,6)
val m = l.grouped(2).map { x => x.head -> x.tail.head }.toMap
// Map(1 -> 2, 3 -> 4, 5 -> 6)
but if list may be of odd length, use headOption:
val l = List(1,2,3,4,5,6,7)
val m = l.grouped(2).map(x => x.head -> x.tail.headOption).toMap
// Map(1 -> Some(2), 3 -> Some(4), 5 -> Some(6), 7 -> None)
Without using grouped that appears ubiquitous in the answers so far.
scala> val l = (1 to 6).toList
l: List[Int] = List(1, 2, 3, 4, 5, 6)
scala> l.zip(l.tail).zipWithIndex.collect { case (e, pos) if pos % 2 == 0 => e }.toMap
res0: scala.collection.immutable.Map[Int,Int] = Map(1 -> 2, 3 -> 4, 5 -> 6)
You may also use sliding and foldLeft as follows:
scala> l.sliding(2,2).foldLeft(Map.empty[Int,Int]){ case (m, List(l, r)) => m + (l -> r) }
res1: scala.collection.immutable.Map[Int,Int] = Map(1 -> 2, 3 -> 4, 5 -> 6)
Is there a function in Scala to compose two maps or is flatMap a sensible approach?
scala> val caps: Map[String, Int] = Map(("A", 1), ("B", 2))
caps: Map[String,Int] = Map(A -> 1, B -> 2)
scala> val lower: Map[Int, String] = Map((1, "a"), (2, "b"))
lower: Map[Int,String] = Map(1 -> a, 2 -> b)
scala> caps.flatMap {
| case (cap, idx) => Map((cap, lower(idx)))
| }
res1: scala.collection.immutable.Map[String,String] = Map(A -> a, B -> b)
Some syntactic sugar would be great!
If you know lower will contain keys for all the values in caps, you can use mapValues:
scala> caps mapValues lower
res0: scala.collection.immutable.Map[String,String] = Map(A -> a, B -> b)
If you don't want or need a new collection, just a mapping, it's a little more idiomatic to use andThen:
scala> val composed = caps andThen lower
composed: PartialFunction[String,String] = <function1>
scala> composed("A")
res1: String = a
This also assumes there aren't values in caps that aren't mapped in lower.
I want to check if a Map doesn't contain empty values. If the value is empty it shouldn't includen in the new Map.
I tried something like:
val newmap = map.map{ entry => if(!entry._2.isEmpty()) Map(entry._1 -> entry._2)}
This does exactly do what I want, but it is not very nice. Is there a better solution?
scala> Map(1 -> List(3, 4), 2 -> Nil, 3 -> List(11))
res2: scala.collection.immutable.Map[Int,List[Int]] = Map(1 -> List(3, 4), 2 -> List(), 3 -> List(11))
scala> res2.filter(_._2.nonEmpty)
res3: scala.collection.immutable.Map[Int,List[Int]] = Map(1 -> List(3, 4), 3 -> List(11))
scala>
You mean empty as in null?
scala> val map = collection.immutable.HashMap[Int, String] (1 -> "a", 2-> "b", 3 -> null)
map: scala.collection.immutable.HashMap[Int,String] = Map(1 -> a, 2 -> b, 3 -> null)
scala> val newmap=map filter (_._2 != null)
newmap: scala.collection.immutable.HashMap[Int,String] = Map(1 -> a, 2 -> b)
EDIT: dang... #missingfaktor beat me to it... :)
Inside a function of mine I construct a result set by filling a new mutable HashMap with data (if there is a better way - I'd appreciate comments). Then I'd like to return the result set as an immutable HashMap. How to derive an immutable from a mutable?
Discussion about returning immutable.Map vs. immutable.HashMap notwithstanding, what about simply using the toMap method:
scala> val m = collection.mutable.HashMap(1 -> 2, 3 -> 4)
m: scala.collection.mutable.HashMap[Int,Int] = Map(3 -> 4, 1 -> 2)
scala> m.toMap
res22: scala.collection.immutable.Map[Int,Int] = Map(3 -> 4, 1 -> 2)
As of 2.9, this uses the method toMap in TraversableOnce, which is implemented as follows:
def toMap[T, U](implicit ev: A <:< (T, U)): immutable.Map[T, U] = {
val b = immutable.Map.newBuilder[T, U]
for (x <- self)
b += x
b.result
}
scala> val m = collection.mutable.HashMap(1->2,3->4)
m: scala.collection.mutable.HashMap[Int,Int] = Map(3 -> 4, 1 -> 2)
scala> collection.immutable.HashMap() ++ m
res1: scala.collection.immutable.Map[Int,Int] = Map(1 -> 2, 3 -> 4)
or
scala> collection.immutable.HashMap(m.toSeq:_*)
res2: scala.collection.immutable.HashMap[Int,Int] = Map(1 -> 2, 3 -> 4)
If you have a map : logMap: Map[String, String]
just need to do : logMap.toMap()