Question

Suppose I在:

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

我需要的是建立一个新的阵列,作为第一批阵列中相应三个(任何编号)要素的组合,从而形成<代码>(123”、“456”、“7”。

1. 成为新鲜事物我写了以下既不简明又不有效的法典:

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
}

这样做有什么办法?

Answer 1

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

或

strings grouped 3 map (_.mkString) toArray

我个人喜欢第一版:

Answer 2

strings grouped 3 map (_.mkString)

或(为了实际获得<代码>Array>后)

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

Answer 3

......或使用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)

提单:你3,向前迈进3。备选案文:

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)

页: 1

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