I want to update Map value which is present in another Map. When I try to update is says 'value update is not a member of Option[scala.collection.immutable.Map[Int,Int]]'.
I tried to convert the value to Map but still, it didn't work for me.
val map = Map("one" -> Map(1 -> 11), "two" -> Map(2 -> 22))
val value = map1.get("one")
value(1) = 100 //value update is not a member of Option[scala.collection.Map[Int,Int]]
There are two mistakes you are making.
Calling get on a Map will return an Option, hence you are not able to set the value.
You are using immutable Map when your operation/purpose is to update the value of some key, for which you need to use mutable map.
Let us try to do the write some snippets to solve these two problems.
scala> val map = Map("one" -> Map(1 -> 11), "two" -> Map(2 -> 22))
map: scala.collection.immutable.Map[String,scala.collection.immutable.Map[Int,Int]] = Map(one -> Map(1 -> 11), two -> Map(2 -> 22))
scala> val valueOption = map.get("one")
valueOption: Option[scala.collection.immutable.Map[Int,Int]] = Some(Map(1 -> 11))
scala> val value = map("one")
value: scala.collection.immutable.Map[Int,Int] = Map(1 -> 11)
scala> value(1) = 100
<console>:13: error: value update is not a member of scala.collection.immutable.Map[Int,Int]
value(1) = 100
You should notice the difference between getting the value using .get and directly using parenthesis. This is a more understandable error and no need to understand Scala magic happening underneath.
Now if you repeat the same statements after importing mutable Map, you will be able to get what you are trying to achieve.
scala> import scala.collection.mutable.Map
import scala.collection.mutable.Map
scala> val map = Map("one" -> Map(1 -> 11), "two" -> Map(2 -> 22))
map: scala.collection.mutable.Map[String,scala.collection.mutable.Map[Int,Int]] = Map(one -> Map(1 -> 11), two -> Map(2 -> 22))
scala> val value = map("one")
value: scala.collection.mutable.Map[Int,Int] = Map(1 -> 11)
scala> value(1) = 100
scala> map
res2: scala.collection.mutable.Map[String,scala.collection.mutable.Map[Int,Int]] = Map(one -> Map(1 -> 100), two -> Map(2 -> 22))
When you created you first map it is already immutable which cannot be changed
scala> val map = Map("one" -> Map(1 -> 11), "two" -> Map(2 -> 22))
map: scala.collection.immutable.Map[String,scala.collection.immutable.Map[Int,Int]] = Map(one -> Map(1 -> 11), two -> Map(2 -> 22))
Your second command is returning an Option of immutable Map again and that can not be updated too.
scala> val value = map.get("one")
value: Option[scala.collection.immutable.Map[Int,Int]] = Some(Map(1 -> 11))
As chunjef suggested, you should be using mutable Map
scala> val map = Map("one" -> scala.collection.mutable.Map(1 -> 11), "two" -> scala.collection.mutable.Map(2 -> 22))
map: scala.collection.immutable.Map[String,scala.collection.mutable.Map[Int,Int]] = Map(one -> Map(1 -> 11), two -> Map(2 -> 22))
Related
I have LinkedHashMaps of type:
val map1 = LinkedHashMap(1 -> 1, 2 -> (1,2), 3 -> (1,2,3))
val map2 = LinkedHashMap(2 -> 2, 3 -> (2,3), 5 -> (2,3,5))
where the integers are nodes's ids of a graph, and the list is the path to that node. I want to implement the case of deleting a node. Suppose I want to delete node 3, I have to do two actions: remove the element with key = 3 in every map, remove the elements which have 3 in their list. How to do it in scala?
If you define you map like you have,
val map1 = LinkedHashMap(1 -> 1, 2 -> (1,2), 3 -> (1,2,3))
You do not have key: Int and value: List[Int] but you have key: Int and values: Any.
scala> val map1 = LinkedHashMap(1 -> 1, 2 -> (1,2), 3 -> (1,2,3))
// map1: scala.collection.mutable.LinkedHashMap[Int,Any] = Map(1 -> 1, 2 -> (1,2), 3 -> (1,2,3))
To match your requirement, you should define your map like following,
scala> val map1 = LinkedHashMap(1 -> List(1), 2 -> List(1,2), 3 -> List(1,2,3))
// map1: scala.collection.mutable.LinkedHashMap[Int,List[Int]] = Map(1 -> List(1), 2 -> List(1, 2), 3 -> List(1, 2, 3))
Now, if you want to delete a node 3,
scala> val map2 = map1.filter({
| case (key, list) => key != 3 && !list.contains(3)
| })
// map2: scala.collection.mutable.LinkedHashMap[Int,List[Int]] = Map(1 -> List(1), 2 -> List(1, 2))
How can I merge maps like below:
val map1 = Map(1 -> "a", 2 -> "b")
val map2 = Map("a" -> "A", "b" -> "B")
After merged.
Merged = Map( 1 -> List("a", "A"), 2 -> List("b", "B"))
Can be List, Set or any other collection who has size attribute.
I'm not sure I understand what are you searching for exactly, but to achieve that for the provided example you could do:
val map1 = Map(1 -> "a", 2 -> "b")
val map2 = Map("a" -> "A", "b" -> "B")
map1.mapValues(value => (value, map2(value)))
However you should be careful to have every value from a as a key in b (I just assumed this happens from the provided example).
Given two maps with value1 as key2
scala> val x = Map(1 -> "a", 2 -> "b")
x: scala.collection.immutable.Map[Int,String] = Map(1 -> a, 2 -> b)
scala> val y = Map("a" -> "A", "b" -> "B")
y: scala.collection.immutable.Map[String,String] = Map(a -> A, b -> B)
Merge as Map(k1 -> List(v1, v2))
scala> val z = x.map { case (k1, v1) => (k1, List(v1, y(v1))) }
z: scala.collection.immutable.Map[Int,List[String]] = Map(1 -> List(a, A), 2 -> List(b, B))
You basically need to get value from first map then lookup the second map, and just create a List out of those (v1, v2).
Try This
scala> val map1 = Map(1 -> "a", 2 -> "b")
map1: scala.collection.immutable.Map[Int,String] = Map(1 -> a, 2 -> b)
scala> val map2 = Map("a" -> "A", "b" -> "B")
map2: scala.collection.immutable.Map[String,String] = Map(a -> A, b -> B)
scala> map1.zip(map2).map(x=>x._1._1 -> List(x._2._1,x._2._2))
res44: scala.collection.immutable.Map[Int,List[String]] = Map(1 -> List(a, A), 2 -> List(b, B))
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 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)
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... :)