Question

如何在Swift中求和NSNumber数组？我使用.reduce时收到此错误。任何其他方式来总结NSNumber数组？

无法使用类型'（Int，_）
的参数列表调用'reduce'

var record = [Record]()
var incomeFilter: [Record] = []

let formatter = NumberFormatter()
formatter.numberStyle = .currencyAccounting
formatter.currencySymbol = "¥"
formatter.maximumFractionDigits = 2
formatter.minimumFractionDigits = 2

incomeFilter = record.filter { $0.recordtype!.contains("Income") && $0.createdAt! == recordItem.createdAt!}
let incomeSumArray = incomeFilter.map{formatter.number(from: $0.amount!) ?? 0.00}.reduce(0, +)
//Cannot invoke 'reduce' with an argument list of type '(Int, _)
print(incomeSumArray)

//result without .reduce
//[9.99, 6.58, 7777.77]

Answer 1

我认为问题是?? 0使用NSNumber类型的可选项。 nil-coalescing时总是需要相同的类型。您必须先将NSNumber转换为数值（double）：

let incomeSum = incomeFilter
    .map {
        formatter.number(from: $0.amount!)?.doubleValue ?? 0
     }
    .reduce(0, +)

或

let incomeSum = incomeFilter
    .map {
        guard
           let amountString = $0.amount,
           let number = formatter.number(from: amountString)
        else {
           return 0
        }

        return number.doubleValue
     }
    .reduce(0, +)

Answer 2

您可以在减少它之前将incomeFilter映射为双精度数组（incomeFilter.map{formatter.number(from: $0.amount!) ?? 0.00}）：

let incomeSumArray = incomeFilter.map {
        formatter.number(from: $0.amount!) ?? 0.00
    }.map {
        $0.doubleValue
    }.reduce(0, +)

如何在Swift中求和NSNumber数组？

2 个答案: