Find a number available in first array with numbers in second. If number not found get the immediate lower.
val a = List(1,2,3,4,5,6,7,8,9)
val b = List(1,5,10)
expected output after comparing a with b
1 --> 1
2 --> 1
3 --> 1
4 --> 1
5 --> 5
6 --> 5
7 --> 5
8 --> 5
9 --> 5
Thanks
You can use TreeSet's to() and lastOption methods as follows:
val a = List(1, 2, 3, 4, 5, 6, 7, 8, 9)
val b = List(1, 5, 10)
import scala.collection.immutable.TreeSet
// Convert list `b` to TreeSet
val bs = TreeSet(b.toSeq: _*)
a.map( x => (x, bs.to(x).lastOption.getOrElse(Int.MinValue)) ).toMap
// res1: scala.collection.immutable.Map[Int,Int] = Map(
// 5 -> 5, 1 -> 1, 6 -> 5, 9 -> 5, 2 -> 1, 7 -> 5, 3 -> 1, 8 -> 5, 4 -> 1
// )
Note that neither list a or b needs to be ordered.
UPDATE:
Starting Scala 2.13, methods to for TreeSet is replaced with rangeTo.
Here is another approach using collect function
val a = List(1,2,3,4,5,6,7,8,9)
val b = List(1,5,10)
val result = a.collect{
case e if(b.filter(_<=e).size>0) => e -> b.filter(_<=e).reverse.head
}
//result: List[(Int, Int)] = List((1,1), (2,1), (3,1), (4,1), (5,5), (6,5), (7,5), (8,5), (9,5))
Here for every element in a check if there is a number in b i.e. which is greater than or equal to it and reverse the filter list and get its head to make it a pair.
apache-spark?map.