is there a way in Scala to sort an array of tuples using and arbitrary comparison function? In particular I need to sort and array of tuples by their second element, but I wanted to know a general technique to sort arrays of tuples.
Thanks!
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7 Answers
In scala 2.8, there is a method sortBy. Here is a simple use case:
scala> val arr = Array(("One",1),("Two",2),("Four",4),("Three",3))
arr: Array[(java.lang.String, Int)] = Array((One,1), (Two,2), (Four,4), (Three,3))
scala> arr.sortBy(_._2)
res0: Array[(java.lang.String, Int)] = Array((One,1), (Two,2), (Three,3), (Four,4))
scala>
4 Comments
tstenner
2.8 is just plain cheating. +1
pau.estalella
Gooood. I'll have to switch to 2.8 (currently using 2.7) Thanks
Daniel C. Sobral
This does not sort in place, however.
Eastsun
@Daniel Yes, and I just wrote a post about it: old.nabble.com/…
You can use this code:
scala> val v = Array(('a', 2), ('b', 1))
v: Array[(Char, Int)] = Array((a,2), (b,1))
scala> scala.util.Sorting.stableSort(v,
| (e1: (Char, Int), e2: (Char, Int)) => e1._2 < e2._2)
scala> v
res11: Array[(Char, Int)] = Array((b,1), (a,2))
Unfortunetly, it seems that Scala cannot infer the type of the array passed to stableSort. I hope that's ok for you.
2 Comments
Erik Kaplun
Bit nicer:
stableSort(v, (_._2 < _._2) : ((Char,Int),(Char,Int)) => Boolean) — keeps the concerns separate, and allows one to reason about the logic and the types as independent steps, especially since the inline type signature is only a nuisance here.
Paul
One things to note is that if you are dealing with the Any type in the sort column (suppose it has Long) you need to do: scala.util.Sorting.stableSort(v, (e1: (String, Any), e2: (String, Any)) => e1._2.asInstanceOf[Long] < e2._2.asInstanceOf[Long])
If it's an Array, it's probably typical to use in-place sorting algorithms. However, in idiomatic Scala code, mutable collections are usually not encouraged/used. If that's the case and you have am immutable collection (or would like to not modify the Array in place), use sortWith:
scala> val a = Array(1, 3, 2, 5)
a: Array[Int] = Array(1, 3, 2, 5)
scala> a.sortWith(_ > _)
res6: Array[Int] = Array(5, 3, 2, 1)
scala> a
res7: Array[Int] = Array(1, 3, 2, 5)
sorting an Array or any other collection of tuples:
scala> val a = Array(('a', 1), ('b', 4), ('c', 5), ('d', 2))
a: Array[(Char, Int)] = Array((a,1), (b,4), (c,5), (d,2))
scala> a.sortWith(_._2 > _._2)
res4: Array[(Char, Int)] = Array((c,5), (b,4), (d,2), (a,1))
scala> a
res5: Array[(Char, Int)] = Array((a,1), (b,4), (c,5), (d,2))
2 Comments
martin jakubik
where can we find about about that "_._2" syntax? What is that called?
Erik Kaplun
Maybe this will help: stackoverflow.com/questions/8000903/…
On Scala 2.8 (yes, again :), you can also do this:
val v = Array(('a', 2), ('b', 1))
scala.util.Sorting.stableSort(v)(manifest[(Char, Int)], Ordering.by(_._2))
In the specific case of pairs, this can also work to sort first by the second element, and then by the first:
scala.util.Sorting.stableSort(v)(manifest[(Char, Int)], Ordering.by(_.swap))
answered Apr 13, 2010 at 12:28
Daniel C. Sobral
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Comments
2.7 and not in place:
(Array((2,3), (4,2), (1,5)).toList.sort (_._2 < _._2)).toArray
Comments
You probably want def stableSort[K](a : Seq[K], f : (K, K) => Boolean) : Array[K] from scala.util.Sorting.
Your comparison function would be something like _._2 < _._1
Comments
val l = List((2, 1), (3, 2), (0, 3))
l sort { case(a, b) => a > b }
1 Comment
pau.estalella
Problem is I have an Array :(