将列表[(字符串，字符串)]转换为Map[String，Map[String，String]]

Question

假设我有下面的列表(字符串，字符串)

List((recap_items[4].invoice_items[0].id,6),
(recap_items[4].invoice_items[1].id,7),
(recap_items[4].invoice_items[1].qty,1),
(recap_items[4].invoice_items[0].qty,1), 
(recap_items[4].invoice_items[1].sur_key,19), 
(recap_items[4].invoice_items[0].sur_key,17))

我如何把这个列表转换成下面的地图呢？

Map(
recap_items[4].invoice_items[0] -> Map(id -> 6, qty -> 1, sur_key -> 17),
recap_items[4].invoice_items[1] -> Map(id -> 7, qty -> 1, sur_key -> 19)
)

或者有什么更好的表现方式来代表这个列表呢？

编辑

case class Recap(recap_id: String, recap_date: Date, submitted_id:String, edited_id: String, recap_items: List[Recap_items])

case class Recap_items(product_name: String, product_id: String, qty: Int, unit_name: String, unit_multiplier: Int, sys_qty: Int, invoice_items: List[Invoice_items])

case class Invoice_items(sur_key: Long, id: Long, qty: Int)

电流逼近

下面是我当前的方法，它给出了Map[String，ListString]：

代码：

flash.data.filterKeys(_.startsWith("recap_items["+i+"].invoice_items")).toList.sortBy(x => x._1).map{
            x => (x._1.split("""\.""").toList(1), x._2)
        }.groupBy(_._1).mapValues{
            x => x.map( v => v._2)
        }

产出：

Map(invoice_items[1] -> List(7, 1, 19), 
invoice_items[0] -> List(6, 1, 17))

知道如何改进这段代码吗？

Answer 1

不确定为什么要将所有内容转换为字符串(我的意思是recap_items4.invoice_items1，而不是实际的项名)。

如果您的输入是Recap对象，则只需执行以下操作：

val recap: Recap = ...
val map: Map[String, Map[String, String]] =
  (for (
    (recapItem, recapItemIndex) <- recap.recap_items.zipWithIndex;
    (invoiceItem, invoiceItemIndex) <- recapItem.invoice_items.zipWithIndex
  ) yield {
    s"recap_item[$recapItemIndex]invoiceItem[$invoiceItemIndex]" -> Map("id" -> s"$invoiceItem.id", "qty" -> s"$invoiceItem.qty", "sur_key" -> s"$invoiceItem.sur_key")
  }).toMap