仅转换Scala列表的第一个元素 [英] Transform only the first element of a Scala list

问题描述

  val head = l1.head 
 val tail = l1.tail 
 val l2 = change（head）:: tail 
  

已更新（）看起来可能会起作用，但没有太大的改善：

  val head = l1.head 
 val l2 = l.update（0，change（head））
  

我喜欢这样：

  val l2 = l1。 updateHead（change（_））
  

有什么吗？

解决方案

您可以尝试使用模式匹配

  val l2 = l1 match {
 case Nil => Nil 
 case x :: xs => change（x）:: xs 
} 
  

这样你就不必担心如果 head 实际返回元素

Is there a way to transform only the first element of a list without doing something super hacky like:

val head = l1.head
val tail = l1.tail
val l2   = change(head) :: tail

updated() looks like it could work, but isn't much of an improvement:

val head = l1.head
val l2   = l.update(0, change(head))

I'd love something like:

val l2   = l1.updateHead(change(_))

Is there anything like that?

解决方案

you could try using pattern matching

val l2 = l1 match{
    case Nil   => Nil
    case x::xs => change(x)::xs
}

That way you don't have to worry if head actually returns an element

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