搜索映射列表以返回更高值

Question

我目前的系统可以找到整数列表的最后一个值，如下面的第一个文件所示。然后将其显示为“当前”值。当我输入一个搜索并返回所有比我搜索的结果更高的结果时，我需要能够找到所有的“当前”值。列表保存为一个Map（String，List [Int]）。

SK1, 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, 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 
SK4, 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, 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, 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, 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, 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, 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 

这就是我想要得到的;使用上面的数据文件，如果我搜索“SK5”，系统将返回;

图高于SK5 - 3

SK4 - 8

SK6 - 9

SK8 - 6

SK9 - 6

这是我的当前的代码以找到列表中的最后一个号码

//5 - Show Current Stock Level (W) 
    def handleFive(): Boolean = { 
    mnuShowSingleDataStock(currentStockLevel) 
    true 
    } 

// Returns a single result, not a list 
    def mnuShowSingleDataStock(stock: (String) => (String,Int)) = { 
    print("Stock > ") 
    val data = stock(readLine) 
    println(s"${data._1}: ${data._2}") 
    } 

    //Show higher than stocks 
    def higherThan(stock: String): List[(String, List[Int])] = { 
    mapdata.toList.sortWith(_._2.last > _._2.last).takeWhile(row => row._2.last > mapdata.get(stock).map(_.last).getOrElse(0)) 
    } 

根据最后的值，然后

Answer 1

排序需要而当前的股票价值是大于搜索股票价值

def higherThan(stock: String): List[(String, List[Int])] = { 
    mapdata.toList.sortWith(_._2.last > _._2.last).takeWhile(row => row._2.last > mapdata.get(stock).map(_.last).getOrElse(0)) 
}