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))
Related
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))
Suppose I have a Map like
val x = Map(1 -> List("a", "b"), 2 -> List("a"),
3 -> List("a", "b"), 4 -> List("a"),
5 -> List("c"))
How would I create from this a new Map where the keys are Lists of keys from x having the same value, e.g., how can I implement
def someFunction(m: Map[Int, List[String]]): Map[List[Int], List[String]] =
// stuff that would turn x into
// Map(List(1, 3) -> List("a", "b"), List(2, 4) -> List("a"), List(5) -> List("c"))
?
You can convert the Map to a List and then use groupBy to aggregate the first element of each tuple:
x.toList.groupBy(_._2).mapValues(_.map(_._1)).map{ case (x, y) => (y, x) }
// res37: scala.collection.immutable.Map[List[Int],List[String]] =
// Map(List(2, 4) -> List(a), List(1, 3) -> List(a, b), List(5) -> List(c))
Or as #Dylan commented, use _.swap to switch the tuples' elements:
x.toList.groupBy(_._2).mapValues(_.map(_._1)).map(_.swap)
I need to reverse a map
customerIdToAccountIds:Map[Int, Seq[Int]]
such that each account ID is a key to a list of all the customer IDs of the account (many-to-many relationship):
accountIdToCustomerIds:Map[Int, Seq[Int]]
What is a good idiomatic way to accomplish this? Thanks!
Input:
val customerIdToAccountIds:Map[Int, Seq[Int]] = Map(1 -> Seq(5,6,7), 2 -> Seq(5,6,7), 3 -> Seq(5,7,8))
val accountIdToCustomerIds:Map[Int, Seq[Int]] = ???
1 -> Seq(5,6,7)
2 -> Seq(5,6,7)
3 -> Seq(5,7,8)
Output:
5 -> Seq(1,2,3)
6 -> Seq(1,2)
7 -> Seq(1,2,3)
8 -> Seq(3)
val m = Map( 1 -> Seq(5,6,7)
, 2 -> Seq(5,6,7)
, 3 -> Seq(5,7,8) )
// Map inverter: from (k -> List(vs)) to (v -> List(ks))
m flatten {case(k, vs) => vs.map((_, k))} groupBy (_._1) mapValues {_.map(_._2)}
//result: Map(8 -> List(3), 5 -> List(1, 2, 3), 7 -> List(1, 2, 3), 6 -> List(1, 2))
val customerIdToAccountIds = Map(1 -> Seq(5, 6, 7), 2 -> Seq(5, 6, 7), 3 -> Seq(5, 7, 8))
val accountIdToCustomerIds = customerIdToAccountIds.toSeq.flatMap {
case (customerId, accountIds) => accountIds.map { accountId => (customerId, accountId) } // swap
}.groupBy(_._2).mapValues(_.map(_._1)) // groupBy accountId and extract customerId from tuples
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)
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... :)