我当前的系统是一个映射的String，List [Int]，该String是键值，“ Sk1”，“ Sk2”等，而int是一个0-9的数字列表。
这是我当前查找所有列表的方法，如何编辑此方法以仅查找大于所选“ SK *”的所有“ Sk *”。列表的值是尾部的最后一个元素，我已经找到了它的功能。这是handleFive选项菜单。为了澄清，我需要找到最后一个元素（已经具有该功能），然后仅显示大于所选库存的库存。
菜单选项的处理程序
def handleFive(): Boolean = {
mnuShowSingleDataStock(currentStockLevel)
true
}
def handleSeven(): Boolean = {
mnuShowPointsForStock(allStockLevel)
true
}
调用用户并与用户交互的功能
// Returns a single result, not a list
def mnuShowSingleDataStock(f: (String) => (String,Int)) = {
print("Stock > ")
val data = f(readLine)
println(s"${data._1}: ${data._2}")
}
//Returns a list value
def mnuShowPointsForStock(f: (String) => (String,List[Int])) = {
print("Stock > ")
val data = f(readLine)
println(s"${data._1}: ${data._2}")
}
不确定如何编辑它，当前它显示列表中的所有值，我只想返回大于所选值的值
//Show last element in the list, most current
def currentStockLevel (stock: String): (String, Int) = {
(stock, mapdata.get (stock).map(findLast(_)).getOrElse(0))
}
//Unsure how to change this to only return values greater than the selected one, not everything
def currentStockLevel (stock: String): (String, List[Int]) = {
(stock, mapdata.get (stock).map(findLast(_)).getOrElse(0))
}
我当前的映射列表-这是MAPDATA
val mapdata = Map(
"SK1" -> List(9, 7, 2, 0, 7, 3, 7, 9, 1, 2, 8, 1, 9, 6, 5, 3, 2, 2, 7, 2, 8, 5, 4, 5, 1, 6, 5, 2, 4, 1),
"SK2" -> List(0, 7, 6, 3, 3, 3, 1, 6, 9, 2, 9, 7, 8, 7, 3, 6, 3, 5, 5, 2, 9, 7, 3, 4, 6, 3, 4, 3, 4, 1),
"SK3" -> List(8, 7, 1, 8, 0, 5, 8, 3, 5, 9, 7, 5, 4, 7, 9, 8, 1, 4, 6, 5, 6, 6, 3, 6, 8, 8, 7, 4, 0, 6),
"SK4" -> List(2, 9, 5, 7, 0, 8, 6, 6, 7, 9, 0, 1, 3, 1, 6, 0, 0, 1, 3, 8, 5, 4, 0, 9, 7, 1, 4, 5, 2, 8),
"SK5" -> List(2, 6, 8, 0, 3, 5, 5, 2, 5, 9, 4, 5, 3, 5, 7, 8, 8, 2, 5, 9, 3, 8, 6, 7, 8, 7, 4, 1, 2, 3),
"SK6" -> List(2, 7, 5, 9, 1, 9, 8, 4, 1, 7, 3, 7, 0, 8, 4, 5, 9, 2, 4, 4, 8, 7, 9, 2, 2, 7, 9, 1, 6, 9),
"SK7" -> List(6, 9, 5, 0, 0, 0, 0, 5, 8, 3, 8, 7, 1, 9, 6, 1, 5, 3, 4, 7, 9, 5, 5, 9, 1, 4, 4, 0, 2, 0),
"SK8" -> List(2, 8, 8, 3, 1, 1, 0, 8, 5, 9, 0, 3, 1, 6, 8, 7, 9, 6, 7, 7, 0, 9, 5, 2, 5, 0, 2, 1, 8, 6),
"SK9" -> List(7, 1, 8, 8, 4, 4, 2, 2, 7, 4, 0, 6, 9, 5, 5, 4, 9, 1, 8, 6, 3, 4, 8, 2, 7, 9, 7, 2, 6, 6)
)
该Map[String, List[Int]]类型具有一种filterKeys(f: String => Boolean)方法，以便仅保留满足给定谓词的键。
一个可能的解决方案是
// get int value from stock if of the form "SK<int>"
def stockInt(stock: String): Option[Int] =
Try(stock.drop(2).toInt).filter(_ => stock.startsWith("SK")).toOption
// we keep the keys in the return, so that you do not get unordered results
// (order is not assured by Map)
def currentStockLevel(stock: String): (String, Map[String, Int]) = {
val maybeN = stockInt(stock)
def isGreater(other: String) = (for {
o <- stockInt(other)
n <- maybeN
} yield o > n).getOrElse(true) // if any key is not in the form of SK*, assume it is greater than the original stock
(
stock,
mapdata.filterKeys(isGreater(_)).mapValues(findLast(_))
)
}
如果您确定只有“ SK”键，则另一种可能的方法是使用SortedMap，它使用aSortedSet作为其键，以便确保按需要将键值对进行排序。
在这种情况下，解决方案是
//put all values in mapdata in a SortedMap
val sortedMap = SortedMap[String, List[Int]]() ++ mapdata
def currentStockLevel(stock: String): (String, List[Int]) = {
(
stock,
sortedMap.dropWhile(_ <= stock).toList.map(_._2).map(findLast(_))
)
}
编辑（在对预期收益的评论之后）：
如果我很了解您要执行的操作，那么您希望过滤值而不是键。这不是问题，Map也有一种filter(p: ((K, V)) => Boolean): Map[K, V]方法可以做到这一点：
def currentHigherStockLevel(stock: String): Map[String, Int] = {
val current = datamap.get(stock).map(findLast).getOrElse(0) // if stock is not in the keySet, we keep all keys, by keeping those greater than 0.
datamap.mapValues(findLast).filter {
case (sk, val) => val > current
}
}
这将返回一个Map[String; Int]，其中值是最后一个大于作为参数给定值的值（我们保留它们的键，因为它们可能会很有用）。
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