Scala List 函数，用于对连续的相同元素进行分组 [英] Scala List function for grouping consecutive identical elements

问题描述

给定例如:

List(5, 2, 3, 3, 3, 5, 5, 3, 3, 2, 2, 2)

我想去:

List(List(5), List(2), List(3, 3, 3), List(5, 5), List(3, 3), List(2, 2, 2))

我认为有一个简单的 List 函数可以做到这一点，但我找不到它.

I would assume there is a simple List function that does this, but am unable to find it.

推荐答案

这是我通常使用的技巧:

This is the trick that I normally use:

def split[T](list: List[T]) : List[List[T]] = list match {
  case Nil => Nil
  case h::t => val segment = list takeWhile {h ==}
    segment :: split(list drop segment.length)
}

实际上......不是，我通常对集合类型进行抽象并使用尾递归进行优化，但希望保持简单的答案.

Actually... It's not, I usually abstract over the collection type and optimize with tail recursion as well, but wanted to keep the answer simple.

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