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[英]Scala Convert a Map[String, List[String]] to a Map[String, String]
[英]Convert List[(String,String)] to Map[String, Map[String,String]]
假設我在List [(String,String)]下面:
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))
我該如何將該列表轉換為Map以下?
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)
)
還是有一些更好的表示形式可以列出該列表?
編輯
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)
當前的方法
下面是我目前的方法,該方法為我提供了Map [String,List [String]]:
碼:
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)
}
輸出:
Map(invoice_items[1] -> List(7, 1, 19),
invoice_items[0] -> List(6, 1, 17))
任何線索如何改進此代碼?
不知道為什么要將所有內容都轉換為字符串(我的意思是recap_items [4] .invoice_items [1]而不是實際的商品名稱)。
如果您的輸入是Recap
對象,則可以執行以下操作:
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
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