Question

在Scala数组中查找每个不同元素的计数的最有效实现是什么？

Answer 1

你可以这样做：

val xs = Array("a", "b", "c", "c", "a", "b", "c", "b", "b", "a")
xs.groupBy(identity).mapValues(_.length)

或者像这样：

xs.foldLeft(Map[String, Int]().withDefaultValue(0))((acc, x) => acc + (x -> (acc(x) + 1)))

如果您想提高效率，可以使用内部可变性来防止复制：

def countElems[A](xs: Array[A]): Map[A, Int] = {
  val result = collection.mutable.Map[A, Int]().withDefaultValue(0)
  xs foreach { x => result += (x -> (result(x) + 1)) }
  result.toMap // this copies and makes it immutable, O(number of distinct elements)
}

查找Scala数组中每个不同元素的计数？

1 个答案: