我可以简化以下zipped和zipWithIndex吗？

Question

def creditTransfer(invoices: List[Invoice], salons: List[Salon],
                   financialDetails: List[FinancialDetail]) = {
    // TODO: can the following be simplified? instead of having a
    // zipped and a zipWithIndex maybe one zip or something ...
    val creditTxnInforation = (invoices, salons, financialDetails)
    .zipped.toList.zipWithIndex.map {
      case ((invoice, salon, financialDetail), index) =>
        InvoiceToSepaConverter.toCreditTransactionDetail(
           invoice, salon, financialDetail, paymentId(pmtInfId, index))
    }
}

正如评论所解释的那样，我想简化zipped和zipWithIndex。

Answer 1

将付款ID转换为可迭代，以便您可以将索引概念推到一边，并一次压缩所有四个迭代。

不确定这是否完全正确，因为我没有足够的代码来编译它，但这些内容是：

def creditTransfer(invoices: List[Invoice], salons: List[Salon],
                   financialDetails: List[FinancialDetail]) = {
  val paymentIds = Stream.from(0).map(paymentId(pmtInfId, _))
  val creditTxnInforation =
    (invoices, salons, financialDetails, paymentIds).zipped
      .map((InvoiceToSepaConverter.toCreditTransactionDetail _).tupled)
}