Question

这是一个简单的代码：

struct Car {
    var model: String
    var producer: String
    var year: Int
}

func AverageYear(years : [Int]) -> Int {
    var x : Int = 0
    for i in 0..<years.count {
        x += years[i]
    }
    return x / years.count
}

var list = [Car(model: "C400", producer: "Mercedes-benz", year: 2014),
            Car(model: "GLE450", producer: "Mercedes-benz", year: 2016)]

print(AverageYear([list[0].year, list[1].year]))

如何在不枚举所有元素的情况下替换AverageYear([list[0].year, list[1].year])？像AverageYear([list.year])

这样的东西

Answer 1

你可以使用 map 从给定的汽车数组构建年数组：

var list = [Car(model: "C400", producer: "Mercedes-benz", year: 2014),
            Car(model: "GLE450", producer: "Mercedes-benz", year: 2016)]

let years = list.map { $0.year }
print(AverageYear(years))

Answer 2

Reduce让它在单行中运行：

struct Car {
  var model: String
  var producer: String
  var year: Int
}

var list = [Car(model: "C400", producer: "Mercedes-benz", year: 2014),
            Car(model: "GLE450", producer: "Mercedes-benz", year: 2016)]

// slice the array to get the elements you want
let slice = list[0...1]

// sum every car.year in the slice and divide by the count
let averageYear: Int = slice.reduce(0){ $0 + $1.year } / slice.count

print(averageYear) // -> 2015

请注意，我使用了list[0...1]，这意味着“元素0到1，包括”。它返回原始数组的切片。如果我有10个元素，我可以list[5...8]，它会给我一个包含元素5,6,7,8的数组切片。

struct arguments - array

2 个答案: