CODEWITHSUNDEEP
scalafor-loop
nab
nab

Reputation: 4861

Proper way for looping in Scala

Suppose I have an array of strings in Scala:

val strings = Array[String]("1", "2", "3", "4", "5", "6", "7")

What I need is to make a new array which elements will be obtained as a concatenation of each three (any number) consequent elements of the first array, which should result in ("123", "456", "7")

Being new to Scala I wrote the following code which was neither concise nor effective:

var step = 3
val strings = Array[String]("1", "2", "3", "4", "5", "6", "7")
val newStrings = collection.mutable.ArrayBuffer.empty[String]

for (i <- 0 until strings.length by step) {
  var elem = ""
  for (k <- 0 until step if i + k < strings.length) {
    elem += strings(i + k)
  }
  newStrings += elem
}

What would be the Scala way for doing this?

Upvotes: 2

Views: 529

Answers (3)

user unknown
user unknown

Reputation: 36259

... or use sliding 

val strings = Array[String]("1", "2", "3", "4", "5", "6", "7")
strings.sliding (3, 3) .map (_.mkString).toArray

res19: Array[String] = Array(123, 456, 7)

Sliding: You take 3, and move forward 3. Variants: 

scala> strings.sliding (3, 2) .map (_.mkString).toArray
res20: Array[String] = Array(123, 345, 567)

take 3, but forward 2

scala> strings.sliding (2, 3) .map (_.mkString).toArray
res21: Array[String] = Array(12, 45, 7)

take 2, forward 3 (thereby skipping every third)

Upvotes: 3

Debilski
Debilski

Reputation: 67888

strings grouped 3 map (_.mkString)

or (in order to really get an Array back)

(strings grouped 3 map (_.mkString)).toArray

Upvotes: 4

Rogach
Rogach

Reputation: 27250

strings.grouped(3).map(_.mkString).toArray

or

strings grouped 3 map (_.mkString) toArray

I personally prefer the first version :)

Upvotes: 9

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