Scala grouped and sliding · GitHub - Gist
gist.github.com › ghiden › 4721613Scala grouped and sliding · GitHub Instantly share code, notes, and snippets. ghiden / scala_grouped_and_sliding Created 10 years ago Star 1 Fork 0 Code Revisions 2 Stars 1 Scala grouped and sliding Raw scala_grouped_and_sliding scala> val a = (1 to 10).toList a: List [Int] = List (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
Scala: grouped from right? - Stack Overflow
stackoverflow.com › questions › 17075607Jun 13, 2013 · In Scala, grouped works from left to right. val list = List(1,2,3,4,5) list.grouped(2).toList => List[List[Int]] = List(List(1, 2), List(3, 4), List(5)) But what if I want: => List[List[Int]] = List(List(1), List(2, 3), List(4, 5)) ? Well I know this works: list.reverse.grouped(2).map(_.reverse).toList.reverse It seems not efficient, however.
Scala Tutorial - GroupBy Function Example
allaboutscala.com › scala-groupby-exampleMar 16, 2018 · Overview. In this tutorial, we will learn how to use the groupBy function with examples on collection data structures in Scala. The groupBy function is applicable to both Scala's Mutable and Immutable collection data structures. The groupBy method takes a predicate function as its parameter and uses it to group elements by key and values into a Map collection.
Scala: sliding(N,N) vs grouped(N) - Stack Overflow
stackoverflow.com › questions › 32874419Scala: sliding (N,N) vs grouped (N) Ask Question. Asked 7 years, 3 months ago. Modified 2 years, 7 months ago. Viewed 27k times. 32. I found myself lately using sliding (n,n) when I need to iterate collections in groups of n elements without re-processing any of them. I was wondering if it would be more correct to iterate those collections by using grouped (n).
Scala: grouped from right? - Stack Overflow
https://stackoverflow.com/questions/17075607In Scala, grouped works from left to right. val list = List(1,2,3,4,5) list.grouped(2).toList => List[List[Int]] = List(List(1, 2), List(3, 4), List(5)) But what if I want: => List[List[Int]] = List(List(1), List(2, 3), List(4, 5)) ? Well I know this works: list.reverse.grouped(2).map(_.reverse).toList.reverse It seems not efficient, however.