第n个元素我有一个List
获取从列表中
val family=List("1","2","11","12","21","22","31","33","41","44","51","55")
我想利用它的第n个元素,但问题是,parents
大小不固定。
val familliar=List("1","2","11") //n=3
第n个元素我有一个List
获取从列表中
val family=List("1","2","11","12","21","22","31","33","41","44","51","55")
我想利用它的第n个元素,但问题是,parents
大小不固定。
val familliar=List("1","2","11") //n=3
您可以使用take
scala> val list = List(1,2,3,4,5,6,7,8,9)
list: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9)
scala> list.take(3)
res0: List[Int] = List(1, 2, 3)
使用take
:
val familliar = family.take(3)
List(1,2,3).take(100) //List(1,2,3)
采取的签名会比较指数的说法,所以增量指标会比从不多论点
拍下的签名
override def take(n: Int): List[A] = {
val b = new ListBuffer[A]
var i = 0
var these = this
while (!these.isEmpty && i < n) {
i += 1
b += these.head
these = these.tail
}
if (these.isEmpty) this
else b.toList
}