CODEWITHSUNDEEP
stringscalasequences
J.Hammond
J.Hammond

Reputation: 270

Scala: how to transform sequence of strings into a sequence of tuples by splitting strings

I'm trying to transform a sequence like the one bellow

val raw: Seq[String] = Seq("timmy barns", "jimmy smith", "mark middle")

into a sequence that would look like this.

val parsed: Seq[(String, String)] = Seq(("timmy", "barns"), ("jimmy", "smith"), ("mark", "middle"))

The best I can come up with is something like this.

val parsed: Seq[(String, String)] = Seq(raw.map(i => i.split(" ")))

Obviously this won't work, can anyone give me suggestions?

Upvotes: 1

Views: 451

Answers (4)

jwvh
jwvh

Reputation: 51271

This will gracefully, but silently, drop all input that doesn't split() into 2 elements.

raw.map(_.split("\\s+")).collect{case Array(a,b) => (a,b)}

Upvotes: 1

Andrey Tyukin
Andrey Tyukin

Reputation: 44918

If you want to combine it with a more general regex matching anyway, this variant might be preferable:

val P = "(\\w+) +(\\w+)".r 
val result = for (P(x, y) <- raw) yield (x, y)

gives:

result: Seq[(String, String)] = List((timmy,barns), (jimmy,smith), (mark,middle))

Upvotes: 1

Hayk Hakobyan
Hayk Hakobyan

Reputation: 309

My version )

raw.map(_.split(" ") match {
    case Array(a, b) => (a, b)
})

Upvotes: 0

zenwraight
zenwraight

Reputation: 2000

May be something like this:-

val raw: Seq[String] = Seq("timmy barns", "jimmy smith", "mark middle")

val splitRaw = raw.map { x =>
  (x.split(" ").head, x.split(" ").last)
}

Hope this helps!

Upvotes: 0

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