How to get a set out of List[Array[String]] in Scala

Question

I am super new to scala , I have the the following code:

val content = List("a b","a b")

gives me : List[String] = List(a b, a b)

val split_data = content.map(testFile => testFile.split("\t"))

gives me : List[Array[String]] = List(Array(a b), Array(a b))

and I want to get a set out of the list, but split_data.toSet gives me: scala.collection.immutable.Set[Array[String]] = Set(Array(a b), Array(a b))

while I would want

scala.collection.immutable.Set[Array[String]] = Set(Array(a b))

How do I do that ?

Wait.. Don't do that.. though it outputs the expected result, but that's not what you want.. — Bla...
– Bla..., Commented Aug 29, 2018 at 2:32

Roberto Bonvallet · Accepted Answer · 2018-08-29 02:45:37Z

3

In Scala, two arrays with the same values are not equal, as opposed to other collections:

Array(1, 2, 3) == Array(1, 2, 3) // false
List(1, 2, 3)  == List(1, 2, 3)  // true

This is because arrays are just Java arrays, and those don't have the same semantics as collections in functional languages. In Java, two array references are equal if they point to the same array in memory:

val a = Array(1, 2, 3)
a == a              // true, it's the same array
a == Array(1, 2, 3) // false, it's two different arrays

So, if you want to have a set of unique collections, you have to convert the arrays to Scala sequences:

val content = List("a b","a b")
val split_data = content.map(testFile => testFile.split("\t").toSeq)  // note the .toSeq
split_data.toSet   // set with one element

edited Aug 29, 2018 at 2:45

answered Aug 29, 2018 at 2:37

Roberto Bonvallet

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Dima · Accepted Answer · 2018-08-29 15:35:54Z

2

You can't compare arrays like that as explained in the other answer. The easiest way around this is dedup (convert to Set) before .map:

content.toSet[String].map(_.split("\t"))

This is also a bit more efficient, because you end up spliting fewer strings.

edited Aug 29, 2018 at 15:35

answered Aug 29, 2018 at 11:19

Dima

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3 Comments

jwvh Over a year ago

error: missing parameter type for expanded function (Scala 2.12.6)

Dima Over a year ago

@jwvh hm yeah ... I have no idea why it does that :/

Dima Over a year ago

Ah, I got it ... that's because toSet takes a type parameter, gotta tell it explicitly, that you want String.

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