Custom ordering in TreeMap - scala

Here are examples that I have been playing with:
import collection.immutable.{TreeSet, TreeMap}
val ts = TreeSet(9, 23, 1, 2)
ts
val tm = TreeMap(3 -> "c", 1 -> "a", 2 -> "b")
tm
// convert a map to a sorted map
val m = Map("98" -> List(4, 12, 14), "001" -> List(22, 11))
val t = TreeMap(m.toSeq: _*)
t // sorted by key
// sort an unsorted map
m.toSeq.sortWith((x, y) => x._2(0) < y._2(0))
// add a unsorted map into a sorted map
val m1 = Map("07" -> List(3, 5, 1), "05" -> List(12, 5, 3))
val t1: TreeMap[String, List[Int]] = t ++ m1
t1 // "001" is the first key
I can use sortWith on a Map to get a custom ordering, what if I want to use a TreeMap that uses a different ordering than the default?

You can't use Map's values to define default ordering of a Map.
TreeMap[A,B]'s constructor accepts an implicit Ordering[A] parameter, so you could do something like this:
// Will sort according to default Int ordering (ascending by numeric value)
scala> val tm = TreeMap(3 -> "c", 1 -> "a", 2 -> "b")
tm: scala.collection.immutable.TreeMap[Int,String] = Map(1 -> a, 2 -> b, 3 -> c)
// A wild implicit appears! (orders descending by numeric value)
scala> implicit val tmOrd = Ordering[Int].on((x:Int) => -x)
tmOrd: scala.math.Ordering[Int] = scala.math.Ordering$$anon$5#1d8e2eea
// Our implicit is implicitly (yeah) used by constructor
scala> val invTm = TreeMap(3 -> "c", 1 -> "a", 2 -> "b")
invTm: scala.collection.immutable.TreeMap[Int,String] = Map(3 -> c, 2 -> b, 1 -> a)
Note that it's safer to limit a scope of implicits like this one. If you can, you should construct an (not-implicit) object and pass it manually, or separate the scope of implicit declaration from the place where other code can be affected by its presence.
The reason behind this is that TreeMap is built on top of a tree that uses keys' values to maintain structure constraints that allow for efficient data reads/writes based on keys, which is the primary purpose of a Map. Ordering on values in a Map simply makes no sense.
Upd.: The complexity of ordering logic doesn't mean anything. According to your comment:
scala> object ComplexOrdering extends Ordering[Int] {
| def compare(a: Int, b: Int) = {
| if(a == 3) -1 else if(a == 2 * b) -1 else if(a == 3 * b) 0 else 1
| }
| }
defined object ComplexOrdering
scala> val tm = TreeMap(3 -> "c", 1 -> "a", 2 -> "b")
tm: scala.collection.immutable.TreeMap[Int,String] = Map(1 -> a, 2 -> b, 3 -> c)
scala> val tm = TreeMap(3 -> "c", 1 -> "a", 2 -> "b")(ComplexOrdering)
tm: scala.collection.immutable.TreeMap[Int,String] = Map(3 -> c, 2 -> b, 1 -> a)

TreeMap is defined as a Map-like type with a specified ordering of its keys. That ordering is given by an implicit parameter to the constructor:
new TreeMap()(implicit ordering: Ordering[A]) // For TreeMap[A,B]
so you can set an alternative ordering on the keys at construction by explicitly providing a custom Ordering[A].
The class does not, however, provide any (direct) means of setting an ordering based on the values. What you have with calling .toSeq.sortWith is about the best you can do as far as I know, short of coding your own collection type.

Related

Most efficient way to create matrix from Map's [duplicate]

val map1 = Map(1 -> 9 , 2 -> 20)
val map2 = Map(1 -> 100, 3 -> 300)
I want to merge them, and sum the values of same keys. So the result will be:
Map(2->20, 1->109, 3->300)
Now I have 2 solutions:
val list = map1.toList ++ map2.toList
val merged = list.groupBy ( _._1) .map { case (k,v) => k -> v.map(_._2).sum }
and
val merged = (map1 /: map2) { case (map, (k,v)) =>
map + ( k -> (v + map.getOrElse(k, 0)) )
}
But I want to know if there are any better solutions.
The shortest answer I know of that uses only the standard library is
map1 ++ map2.map{ case (k,v) => k -> (v + map1.getOrElse(k,0)) }
Scalaz has the concept of a Semigroup which captures what you want to do here, and leads to arguably the shortest/cleanest solution:
scala> import scalaz._
import scalaz._
scala> import Scalaz._
import Scalaz._
scala> val map1 = Map(1 -> 9 , 2 -> 20)
map1: scala.collection.immutable.Map[Int,Int] = Map(1 -> 9, 2 -> 20)
scala> val map2 = Map(1 -> 100, 3 -> 300)
map2: scala.collection.immutable.Map[Int,Int] = Map(1 -> 100, 3 -> 300)
scala> map1 |+| map2
res2: scala.collection.immutable.Map[Int,Int] = Map(1 -> 109, 3 -> 300, 2 -> 20)
Specifically, the binary operator for Map[K, V] combines the keys of the maps, folding V's semigroup operator over any duplicate values. The standard semigroup for Int uses the addition operator, so you get the sum of values for each duplicate key.
Edit: A little more detail, as per user482745's request.
Mathematically a semigroup is just a set of values, together with an operator that takes two values from that set, and produces another value from that set. So integers under addition are a semigroup, for example - the + operator combines two ints to make another int.
You can also define a semigroup over the set of "all maps with a given key type and value type", so long as you can come up with some operation that combines two maps to produce a new one which is somehow the combination of the two inputs.
If there are no keys that appear in both maps, this is trivial. If the same key exists in both maps, then we need to combine the two values that the key maps to. Hmm, haven't we just described an operator which combines two entities of the same type? This is why in Scalaz a semigroup for Map[K, V] exists if and only if a Semigroup for V exists - V's semigroup is used to combine the values from two maps which are assigned to the same key.
So because Int is the value type here, the "collision" on the 1 key is resolved by integer addition of the two mapped values (as that's what Int's semigroup operator does), hence 100 + 9. If the values had been Strings, a collision would have resulted in string concatenation of the two mapped values (again, because that's what the semigroup operator for String does).
(And interestingly, because string concatenation is not commutative - that is, "a" + "b" != "b" + "a" - the resulting semigroup operation isn't either. So map1 |+| map2 is different from map2 |+| map1 in the String case, but not in the Int case.)
Quick solution:
(map1.keySet ++ map2.keySet).map {i=> (i,map1.getOrElse(i,0) + map2.getOrElse(i,0))}.toMap
Well, now in scala library (at least in 2.10) there is something you wanted - merged function. BUT it's presented only in HashMap not in Map. It's somewhat confusing. Also the signature is cumbersome - can't imagine why I'd need a key twice and when I'd need to produce a pair with another key. But nevertheless, it works and much cleaner than previous "native" solutions.
val map1 = collection.immutable.HashMap(1 -> 11 , 2 -> 12)
val map2 = collection.immutable.HashMap(1 -> 11 , 2 -> 12)
map1.merged(map2)({ case ((k,v1),(_,v2)) => (k,v1+v2) })
Also in scaladoc mentioned that
The merged method is on average more performant than doing a
traversal and reconstructing a new immutable hash map from
scratch, or ++.
This can be implemented as a Monoid with just plain Scala. Here is a sample implementation. With this approach, we can merge not just 2, but a list of maps.
// Monoid trait
trait Monoid[M] {
def zero: M
def op(a: M, b: M): M
}
The Map based implementation of the Monoid trait that merges two maps.
val mapMonoid = new Monoid[Map[Int, Int]] {
override def zero: Map[Int, Int] = Map()
override def op(a: Map[Int, Int], b: Map[Int, Int]): Map[Int, Int] =
(a.keySet ++ b.keySet) map { k =>
(k, a.getOrElse(k, 0) + b.getOrElse(k, 0))
} toMap
}
Now, if you have a list of maps that needs to be merged (in this case, only 2), it can be done like below.
val map1 = Map(1 -> 9 , 2 -> 20)
val map2 = Map(1 -> 100, 3 -> 300)
val maps = List(map1, map2) // The list can have more maps.
val merged = maps.foldLeft(mapMonoid.zero)(mapMonoid.op)
map1 ++ ( for ( (k,v) <- map2 ) yield ( k -> ( v + map1.getOrElse(k,0) ) ) )
I wrote a blog post about this , check it out :
http://www.nimrodstech.com/scala-map-merge/
basically using scalaz semi group you can achieve this pretty easily
would look something like :
import scalaz.Scalaz._
map1 |+| map2
You can also do that with Cats.
import cats.implicits._
val map1 = Map(1 -> 9 , 2 -> 20)
val map2 = Map(1 -> 100, 3 -> 300)
map1 combine map2 // Map(2 -> 20, 1 -> 109, 3 -> 300)
Starting Scala 2.13, another solution only based on the standard library consists in replacing the groupBy part of your solution with groupMapReduce which (as its name suggests) is an equivalent of a groupBy followed by mapValues and a reduce step:
// val map1 = Map(1 -> 9, 2 -> 20)
// val map2 = Map(1 -> 100, 3 -> 300)
(map1.toSeq ++ map2).groupMapReduce(_._1)(_._2)(_+_)
// Map[Int,Int] = Map(2 -> 20, 1 -> 109, 3 -> 300)
This:
Concatenates the two maps as a sequence of tuples (List((1,9), (2,20), (1,100), (3,300))). For conciseness, map2 is implicitly converted to Seq to adapt to the type of map1.toSeq - but you could choose to make it explicit by using map2.toSeq,
groups elements based on their first tuple part (group part of groupMapReduce),
maps grouped values to their second tuple part (map part of groupMapReduce),
reduces mapped values (_+_) by summing them (reduce part of groupMapReduce).
Andrzej Doyle's answer contains a great explanation of semigroups which allows you to use the |+| operator to join two maps and sum the values for matching keys.
There are many ways something can be defined to be an instance of a typeclass, and unlike the OP you might not want to sum your keys specifically. Or, you might want to do operate on a union rather than an intersection. Scalaz also adds extra functions to Map for this purpose:
https://oss.sonatype.org/service/local/repositories/snapshots/archive/org/scalaz/scalaz_2.11/7.3.0-SNAPSHOT/scalaz_2.11-7.3.0-SNAPSHOT-javadoc.jar/!/index.html#scalaz.std.MapFunctions
You can do
import scalaz.Scalaz._
map1 |+| map2 // As per other answers
map1.intersectWith(map2)(_ + _) // Do things other than sum the values
The fastest and simplest way:
val m1 = Map(1 -> 1.0, 3 -> 3.0, 5 -> 5.2)
val m2 = Map(0 -> 10.0, 3 -> 3.0)
val merged = (m2 foldLeft m1) (
(acc, v) => acc + (v._1 -> (v._2 + acc.getOrElse(v._1, 0.0)))
)
By this way, each of element's immediately added to map.
The second ++ way is:
map1 ++ map2.map { case (k,v) => k -> (v + map1.getOrElse(k,0)) }
Unlike the first way, In a second way for each element in a second map a new List will be created and concatenated to the previous map.
The case expression implicitly creates a new List using unapply method.
Here's what I ended up using:
(a.toSeq ++ b.toSeq).groupBy(_._1).mapValues(_.map(_._2).sum)
This is what I came up with...
def mergeMap(m1: Map[Char, Int], m2: Map[Char, Int]): Map[Char, Int] = {
var map : Map[Char, Int] = Map[Char, Int]() ++ m1
for(p <- m2) {
map = map + (p._1 -> (p._2 + map.getOrElse(p._1,0)))
}
map
}
Using the typeclass pattern, we can merge any Numeric type:
object MapSyntax {
implicit class MapOps[A, B](a: Map[A, B]) {
def plus(b: Map[A, B])(implicit num: Numeric[B]): Map[A, B] = {
b ++ a.map { case (key, value) => key -> num.plus(value, b.getOrElse(key, num.zero)) }
}
}
}
Usage:
import MapSyntax.MapOps
map1 plus map2
Merging a sequence of maps:
maps.reduce(_ plus _)
I've got a small function to do the job, it's in my small library for some frequently used functionality which isn't in standard lib.
It should work for all types of maps, mutable and immutable, not only HashMaps
Here is the usage
scala> import com.daodecode.scalax.collection.extensions._
scala> val merged = Map("1" -> 1, "2" -> 2).mergedWith(Map("1" -> 1, "2" -> 2))(_ + _)
merged: scala.collection.immutable.Map[String,Int] = Map(1 -> 2, 2 -> 4)
https://github.com/jozic/scalax-collection/blob/master/README.md#mergedwith
And here's the body
def mergedWith(another: Map[K, V])(f: (V, V) => V): Repr =
if (another.isEmpty) mapLike.asInstanceOf[Repr]
else {
val mapBuilder = new mutable.MapBuilder[K, V, Repr](mapLike.asInstanceOf[Repr])
another.foreach { case (k, v) =>
mapLike.get(k) match {
case Some(ev) => mapBuilder += k -> f(ev, v)
case _ => mapBuilder += k -> v
}
}
mapBuilder.result()
}
https://github.com/jozic/scalax-collection/blob/master/src%2Fmain%2Fscala%2Fcom%2Fdaodecode%2Fscalax%2Fcollection%2Fextensions%2Fpackage.scala#L190
For anyone coming across an AnyVal error, convert the values as follows.
Error:
"could not find implicit value for parameter num: Numeric[AnyVal]"
(m1.toSeq ++ m2.toSeq).groupBy(_._1).mapValues(_.map(_._2.asInstanceOf[Number].intValue()).sum)

Scala: How to "map" an Array[Int] to a Map[String, Int] using the "map" method?

I have the following Array[Int]: val array = Array(1, 2, 3), for which I have the following mapping relation between an Int and a String:
val a1 = array.map{
case 1 => "A"
case 2 => "B"
case 3 => "C"
}
To create a Map to contain the above mapping relation, I am aware that I can use a foldLeft method:
val a2 = array.foldLeft(Map[String, Int]()) { (m, e) =>
m + (e match {
case 1 => ("A", 1)
case 2 => "B" -> 2
case 3 => "C" -> 3
})
}
which outputs:
a2: scala.collection.immutable.Map[String,Int] = Map(A -> 1, B -> 2, C
-> 3)
This is the result I want. But can I achieve the same result via the map method?
The following codes do not work:
val a3 = array.map[(String, Int), Map[String, Int]] {
case 1 => ("A", 1)
case 2 => ("B", 2)
case 3 => ("C", 3)
}
The signature of map is
def map[B, That](f: A => B)
(implicit bf: CanBuildFrom[Repr, B, That]): That
What is this CanBuildFrom[Repr, B, That]? I tried to read Tribulations of CanBuildFrom but don't really understand it. That article mentioned Scala 2.12+ has provided two implementations for map. But how come I didn't find it when I use Scala 2.12.4?
I mostly use Scala 2.11.12.
Call toMap in the end of your expression:
val a3 = array.map {
case 1 => ("A", 1)
case 2 => ("B", 2)
case 3 => ("C", 3)
}.toMap
I'll first define your function here for the sake of brevity in later explanation:
// worth noting that this function is effectively partial
// i.e. will throw a `MatchError` if n is not in (1, 2, 3)
def toPairs(n: Int): (String, Int) =
n match {
case 1 => "a" -> 1
case 2 => "b" -> 2
case 3 => "c" -> 3
}
One possible way to go (as already highlighted in another answer) is to use toMap, which only works on collection of pairs:
val ns = Array(1, 2, 3)
ns.toMap // doesn't compile
ns.map(toPairs).toMap // does what you want
It is worth noting however that unless you are working with a lazy representation (like an Iterator or a Stream) this will result in two passes over the collection and the creation of unnecessary intermediate collections: the first time by mapping toPairs over the collection and then by turning the whole collection from a collection of pairs to a Map (with toMap).
You can see it clearly in the implementation of toMap.
As suggested in the read you already linked in the answer (and in particular here) You can avoid this double pass in two ways:
you can leverage scala.collection.breakOut, an implementation of CanBuildFrom that you can give map (among others) to change the target collection, provided that you explicitly provide a type hint for the compiler:
val resultMap: Map[String, Int] = ns.map(toPairs)(collection.breakOut)
val resultSet: Set[(String, Int)] = ns.map(toPairs)(collection.breakOut)
otherwise, you can create a view over your collection, which puts it in the lazy wrapper that you need for the operation to not result in a double pass
ns.view.map(toPairs).toMap
You can read more about implicit builder providers and views in this Q&A.
Basically toMap (credits to Sergey Lagutin) is the right answer.
You could actually make the code a bit more compact though:
val a1 = array.map { i => ((i + 64).toChar, i) }.toMap
If you run this code:
val array = Array(1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 0)
val a1 = array.map { i => ((i + 64).toChar, i) }.toMap
println(a1)
You will see this on the console:
Map(E -> 5, J -> 10, F -> 6, A -> 1, # -> 0, G -> 7, L -> 12, B -> 2, C -> 3, H -> 8, K -> 11, D -> 4)

Compare two Maps in Scala

Is there any pre-defined function that I can use to compare two Maps based on the key and give me the difference? Right now, I iterate Map1 and foreach key, I check if there is an element in Map2 and I pattern match to find the difference. Is there a much elegant way to do this?
Consider the difference between the maps converted into sets of tuples,
(m1.toSet diff m2.toSet).toMap
Try:
val diff = (m1.keySet -- m2.keySet) ++ (m2.keySet -- m1.keySet)
diff contains the elements that are in m1 and not in m2 and that are in m2 and not in m1.
This solution looks like right way:
scala> val x = Map(1 -> "a", 2 -> "b", 3 -> "c")
x: scala.collection.immutable.Map[Int,String] = Map(1 -> a, 2 -> b, 3 -> c)
scala> val y = Map(1 -> "a", 2 -> "b", 4 -> "d")
y: scala.collection.immutable.Map[Int,String] = Map(1 -> a, 2 -> b, 4 -> d)
scala> val diff : Map[Int, String] = x -- y.keySet
diff: Map[Int,String] = Map(3 -> c)
Found it here https://gist.github.com/frgomes/69068062e7849dfe9d5a53bd3543fb81
I think the -- operator will do what you're looking for: http://www.scala-lang.org/api/current/index.html#scala.collection.Map#--(xs:scala.collection.GenTraversableOnce[A]):Repr
Although this will probably only work given the assumption that Map2 is always a subset of Map1...

Multi-key Map in Scala

How can I create a Map in Scala which does not only take a single parameter as key, but rather two or three.
val map = //..?
map("abc", 1) = 1
println(map("abc", 2)) // => null
println(map("abc", 1)) // => 1
I tried using tuples as a key, but then I have to assign values like this
map(("abc", 1)) = 1
Can I somehow get rid of the inner parentheses?
You could also use
map += ("abc", 1) -> 1
If the map key represents something (e.g. user info) and if you want to add clarity to your code (especially if you have 3 elements in the key), I would go with a case class as the key. Case classes have equals and hashcode implemented so you can safely use them as keys in a map. The code would be more verbose though:
case class MapKey(s: String, i: Int, d: Double)
val map = Map[MapKey, X](MapKey("a", 1, 1.1) -> "value1", MapKey("b", 2, 2.2) -> "value2")
val map2 = map + (MapKey("c", 3, 3.3) -> "value3")
//or for mutable map
map(MapKey("d", 4, 4.4)) = "value4"
//or
map += MapKey("e", 5, 5.5) -> "value5"
You can add your own enhancement to Map that will do the trick:
import collection.mutable.Map
implicit class EnhancedMap[A,B,C](m: Map[(A,B),C]) {
def update(a: A, b: B, c: C) { m((a,b)) = c }
}
then
val map = Map(("abc", 1) -> 0)
map("abc", 1) = 1
works just fine.
You can use -> syntax for tuples:
map("abc" -> 1) = 1
I got compiling errors using Luigi Plinge's approach. The following approach works for me, which is simpler.
scala> var b = Map[(Int, Int), Int]()
b: scala.collection.mutable.Map[(Int, Int),Int] = Map()
scala> b = b + ((1,1)->2)
b: scala.collection.mutable.Map[(Int, Int),Int] = Map((1,1) -> 2)
scala> b
res15: scala.collection.mutable.Map[(Int, Int),Int] = Map((1,1) -> 2)
scala> b = b + ((1,2)->2)
b: scala.collection.mutable.Map[(Int, Int),Int] = Map((1,1) -> 2, (1,2) -> 2)
scala> b(1,1)
res16: Int = 2
scala> b(1,2)
res17: Int = 2

Best way to merge two maps and sum the values of same key?

val map1 = Map(1 -> 9 , 2 -> 20)
val map2 = Map(1 -> 100, 3 -> 300)
I want to merge them, and sum the values of same keys. So the result will be:
Map(2->20, 1->109, 3->300)
Now I have 2 solutions:
val list = map1.toList ++ map2.toList
val merged = list.groupBy ( _._1) .map { case (k,v) => k -> v.map(_._2).sum }
and
val merged = (map1 /: map2) { case (map, (k,v)) =>
map + ( k -> (v + map.getOrElse(k, 0)) )
}
But I want to know if there are any better solutions.
The shortest answer I know of that uses only the standard library is
map1 ++ map2.map{ case (k,v) => k -> (v + map1.getOrElse(k,0)) }
Scalaz has the concept of a Semigroup which captures what you want to do here, and leads to arguably the shortest/cleanest solution:
scala> import scalaz._
import scalaz._
scala> import Scalaz._
import Scalaz._
scala> val map1 = Map(1 -> 9 , 2 -> 20)
map1: scala.collection.immutable.Map[Int,Int] = Map(1 -> 9, 2 -> 20)
scala> val map2 = Map(1 -> 100, 3 -> 300)
map2: scala.collection.immutable.Map[Int,Int] = Map(1 -> 100, 3 -> 300)
scala> map1 |+| map2
res2: scala.collection.immutable.Map[Int,Int] = Map(1 -> 109, 3 -> 300, 2 -> 20)
Specifically, the binary operator for Map[K, V] combines the keys of the maps, folding V's semigroup operator over any duplicate values. The standard semigroup for Int uses the addition operator, so you get the sum of values for each duplicate key.
Edit: A little more detail, as per user482745's request.
Mathematically a semigroup is just a set of values, together with an operator that takes two values from that set, and produces another value from that set. So integers under addition are a semigroup, for example - the + operator combines two ints to make another int.
You can also define a semigroup over the set of "all maps with a given key type and value type", so long as you can come up with some operation that combines two maps to produce a new one which is somehow the combination of the two inputs.
If there are no keys that appear in both maps, this is trivial. If the same key exists in both maps, then we need to combine the two values that the key maps to. Hmm, haven't we just described an operator which combines two entities of the same type? This is why in Scalaz a semigroup for Map[K, V] exists if and only if a Semigroup for V exists - V's semigroup is used to combine the values from two maps which are assigned to the same key.
So because Int is the value type here, the "collision" on the 1 key is resolved by integer addition of the two mapped values (as that's what Int's semigroup operator does), hence 100 + 9. If the values had been Strings, a collision would have resulted in string concatenation of the two mapped values (again, because that's what the semigroup operator for String does).
(And interestingly, because string concatenation is not commutative - that is, "a" + "b" != "b" + "a" - the resulting semigroup operation isn't either. So map1 |+| map2 is different from map2 |+| map1 in the String case, but not in the Int case.)
Quick solution:
(map1.keySet ++ map2.keySet).map {i=> (i,map1.getOrElse(i,0) + map2.getOrElse(i,0))}.toMap
Well, now in scala library (at least in 2.10) there is something you wanted - merged function. BUT it's presented only in HashMap not in Map. It's somewhat confusing. Also the signature is cumbersome - can't imagine why I'd need a key twice and when I'd need to produce a pair with another key. But nevertheless, it works and much cleaner than previous "native" solutions.
val map1 = collection.immutable.HashMap(1 -> 11 , 2 -> 12)
val map2 = collection.immutable.HashMap(1 -> 11 , 2 -> 12)
map1.merged(map2)({ case ((k,v1),(_,v2)) => (k,v1+v2) })
Also in scaladoc mentioned that
The merged method is on average more performant than doing a
traversal and reconstructing a new immutable hash map from
scratch, or ++.
This can be implemented as a Monoid with just plain Scala. Here is a sample implementation. With this approach, we can merge not just 2, but a list of maps.
// Monoid trait
trait Monoid[M] {
def zero: M
def op(a: M, b: M): M
}
The Map based implementation of the Monoid trait that merges two maps.
val mapMonoid = new Monoid[Map[Int, Int]] {
override def zero: Map[Int, Int] = Map()
override def op(a: Map[Int, Int], b: Map[Int, Int]): Map[Int, Int] =
(a.keySet ++ b.keySet) map { k =>
(k, a.getOrElse(k, 0) + b.getOrElse(k, 0))
} toMap
}
Now, if you have a list of maps that needs to be merged (in this case, only 2), it can be done like below.
val map1 = Map(1 -> 9 , 2 -> 20)
val map2 = Map(1 -> 100, 3 -> 300)
val maps = List(map1, map2) // The list can have more maps.
val merged = maps.foldLeft(mapMonoid.zero)(mapMonoid.op)
map1 ++ ( for ( (k,v) <- map2 ) yield ( k -> ( v + map1.getOrElse(k,0) ) ) )
I wrote a blog post about this , check it out :
http://www.nimrodstech.com/scala-map-merge/
basically using scalaz semi group you can achieve this pretty easily
would look something like :
import scalaz.Scalaz._
map1 |+| map2
You can also do that with Cats.
import cats.implicits._
val map1 = Map(1 -> 9 , 2 -> 20)
val map2 = Map(1 -> 100, 3 -> 300)
map1 combine map2 // Map(2 -> 20, 1 -> 109, 3 -> 300)
Starting Scala 2.13, another solution only based on the standard library consists in replacing the groupBy part of your solution with groupMapReduce which (as its name suggests) is an equivalent of a groupBy followed by mapValues and a reduce step:
// val map1 = Map(1 -> 9, 2 -> 20)
// val map2 = Map(1 -> 100, 3 -> 300)
(map1.toSeq ++ map2).groupMapReduce(_._1)(_._2)(_+_)
// Map[Int,Int] = Map(2 -> 20, 1 -> 109, 3 -> 300)
This:
Concatenates the two maps as a sequence of tuples (List((1,9), (2,20), (1,100), (3,300))). For conciseness, map2 is implicitly converted to Seq to adapt to the type of map1.toSeq - but you could choose to make it explicit by using map2.toSeq,
groups elements based on their first tuple part (group part of groupMapReduce),
maps grouped values to their second tuple part (map part of groupMapReduce),
reduces mapped values (_+_) by summing them (reduce part of groupMapReduce).
Andrzej Doyle's answer contains a great explanation of semigroups which allows you to use the |+| operator to join two maps and sum the values for matching keys.
There are many ways something can be defined to be an instance of a typeclass, and unlike the OP you might not want to sum your keys specifically. Or, you might want to do operate on a union rather than an intersection. Scalaz also adds extra functions to Map for this purpose:
https://oss.sonatype.org/service/local/repositories/snapshots/archive/org/scalaz/scalaz_2.11/7.3.0-SNAPSHOT/scalaz_2.11-7.3.0-SNAPSHOT-javadoc.jar/!/index.html#scalaz.std.MapFunctions
You can do
import scalaz.Scalaz._
map1 |+| map2 // As per other answers
map1.intersectWith(map2)(_ + _) // Do things other than sum the values
The fastest and simplest way:
val m1 = Map(1 -> 1.0, 3 -> 3.0, 5 -> 5.2)
val m2 = Map(0 -> 10.0, 3 -> 3.0)
val merged = (m2 foldLeft m1) (
(acc, v) => acc + (v._1 -> (v._2 + acc.getOrElse(v._1, 0.0)))
)
By this way, each of element's immediately added to map.
The second ++ way is:
map1 ++ map2.map { case (k,v) => k -> (v + map1.getOrElse(k,0)) }
Unlike the first way, In a second way for each element in a second map a new List will be created and concatenated to the previous map.
The case expression implicitly creates a new List using unapply method.
Here's what I ended up using:
(a.toSeq ++ b.toSeq).groupBy(_._1).mapValues(_.map(_._2).sum)
This is what I came up with...
def mergeMap(m1: Map[Char, Int], m2: Map[Char, Int]): Map[Char, Int] = {
var map : Map[Char, Int] = Map[Char, Int]() ++ m1
for(p <- m2) {
map = map + (p._1 -> (p._2 + map.getOrElse(p._1,0)))
}
map
}
Using the typeclass pattern, we can merge any Numeric type:
object MapSyntax {
implicit class MapOps[A, B](a: Map[A, B]) {
def plus(b: Map[A, B])(implicit num: Numeric[B]): Map[A, B] = {
b ++ a.map { case (key, value) => key -> num.plus(value, b.getOrElse(key, num.zero)) }
}
}
}
Usage:
import MapSyntax.MapOps
map1 plus map2
Merging a sequence of maps:
maps.reduce(_ plus _)
I've got a small function to do the job, it's in my small library for some frequently used functionality which isn't in standard lib.
It should work for all types of maps, mutable and immutable, not only HashMaps
Here is the usage
scala> import com.daodecode.scalax.collection.extensions._
scala> val merged = Map("1" -> 1, "2" -> 2).mergedWith(Map("1" -> 1, "2" -> 2))(_ + _)
merged: scala.collection.immutable.Map[String,Int] = Map(1 -> 2, 2 -> 4)
https://github.com/jozic/scalax-collection/blob/master/README.md#mergedwith
And here's the body
def mergedWith(another: Map[K, V])(f: (V, V) => V): Repr =
if (another.isEmpty) mapLike.asInstanceOf[Repr]
else {
val mapBuilder = new mutable.MapBuilder[K, V, Repr](mapLike.asInstanceOf[Repr])
another.foreach { case (k, v) =>
mapLike.get(k) match {
case Some(ev) => mapBuilder += k -> f(ev, v)
case _ => mapBuilder += k -> v
}
}
mapBuilder.result()
}
https://github.com/jozic/scalax-collection/blob/master/src%2Fmain%2Fscala%2Fcom%2Fdaodecode%2Fscalax%2Fcollection%2Fextensions%2Fpackage.scala#L190
For anyone coming across an AnyVal error, convert the values as follows.
Error:
"could not find implicit value for parameter num: Numeric[AnyVal]"
(m1.toSeq ++ m2.toSeq).groupBy(_._1).mapValues(_.map(_._2.asInstanceOf[Number].intValue()).sum)