list 从 Scala 列表中获取头项和尾项
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Get head item and tail items from scala list
提问by Falmarri
Is there a method in scala to get the (single) head element of a List or Seq and the (collection) tail of the list? I know there's
scala 中是否有一种方法可以获取 List 或 Seq 的（单个）头部元素以及列表的（集合）尾部？我知道有
def splitAt(n: Int): (List[A], List[A])
and I can easily grab the single item from the first list of the tuple. But is there any built in method that is basically this?
我可以轻松地从元组的第一个列表中获取单个项目。但是是否有任何内置方法基本上是这样的？
def splitAtHead: (Option[A], List[A])
Like I said, you can easily chain splitAtto return the right signature, but I figured a built in method might be able to save an intermediate tuple.
就像我说的，您可以轻松地通过链接splitAt返回正确的签名，但我认为内置方法可能能够保存中间元组。
Edit:
编辑：
@om-nom-nom's answer is correct, but this is why I couldn't use his 2nd version.
@om-nom-nom 的回答是正确的，但这就是我不能使用他的第二个版本的原因。
List[S](s1, s2, s3, s4).sortBy { _.f (h) } match {
case hd :: tail => recurse(tail)
}
回答by om-nom-nom
You can use pattern matching:
您可以使用模式匹配：
val hd::tail = List(1,2,3,4,5)
//hd: Int = 1
//tail: List[Int] = List(2, 3, 4, 5)
Or just .head/.tail methods:
或者只是 .head/.tail 方法：
val hd = foo.head
// hd: Int = 1
val hdOpt = foo.headOption
// hd: Option[Int] = Some(1)
val tl = foo.tail
// tl: List[Int] = List(2, 3, 4)
回答by Saurav Sahu
The tailmethod returns a collectionconsisting of all elements except the first one (which is basically the head).
该tail方法返回一个由除第一个元素（基本上是）以外的所有元素组成的集合head。
+------------------+------------------------+-------------------------------+
| Input | head | tail |
+------------------+------------------------+-------------------------------+
| List() | NoSuchElementException | UnsupportedOperationException |
| List(1) | 1 | List() |
| List(1, 2, 3, 4) | 1 | List(2, 3, 4) |
| "" | NoSuchElementException | UnsupportedOperationException |
| "A" | 'A' (char) | "" |
| "Hello" | 'H' | "ello" |
+------------------+------------------------+-------------------------------+
Note that the two methods apply to Stringtype as well.
请注意，这两种方法也适用于String类型。
Answering @Leandro question: Yes we can do that, as shown below:
回答@Leandro 问题：是的，我们可以这样做，如下所示：
scala> var a::b::c = List("123", "foo", 2020, "bar")
a: Any = 123
b: Any = foo
c: List[Any] = List(2020, bar)
scala> var a::b::c = List("123", "foo", "bar")
a: String = 123
b: String = foo
c: List[String] = List(bar)
scala> var a::b::c = List("123", "foo")
a: String = 123
b: String = foo
c: List[String] = List()
