我写了一个scala函数,将时间戳列表转换为间隔
def toIntervals(timestamps: List[String]) = {
def helper(timestamps: List[String], accu: List[Long]): List[Long] = {
if (timestamps.tail.isEmpty) accu.reverse
else {
val first = timestamps.head.toLong
val second = timestamps.tail.head.toLong
val newHead = second - first
helper(timestamps.tail, newHead :: accu)
}
}
helper(timestamps, List())
}
而且没有尾声
def toIntervals(timestamps: List[String]) : List[Long] = {
if (timestamps.tail.isEmpty) List()
else {
val first = timestamps.head.toLong
val second = timestamps.tail.head.toLong
val newHead = second - first
newHead :: toIntervals(timestamps.tail)
}
}
但我感觉它有一个/两个衬里,例如map2。有什么建议吗?
最佳答案
(timestamps.tail, timestamps).zipped.map(_.toLong - _.toLong)
是你的单线;但是,仅一次
val times = timestamps.map(_.toLong)
效率更高(这将使其成为两线)。关于scala - 将时间戳转换为间隔,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/16547686/