基于上一个元素的每个值的百分比差异创建数组

这可以很方便地用 zip() 数数

let array: [Double] = [4, 5, 2, 4, 3, 6]

let percentages = [0] + zip(array, array.dropFirst()).map {
    (old, new) in
    return 100.0 * (new - old) / old
}

print(percentages) // [0.0, 25.0, -60.0, 100.0, -25.0, 100.0]

(假设所有数组元素都不为零。)

let percentages = [0] + zip(array, array.dropFirst()).map {
    100.0 * ($1 - $0) / $0
}

你可以这样做:

func percentageDifference(num1: Double, num2: Double) -> Double {
    return (num2 - num1) / num1 * 100
}

var array: [Double] = [4, 5, 2, 4, 3, 6]
var resultArray: [Double] = [0]

for i in (1..<array.count) {
    resultArray.append(percentageDifference(num1: array[i-1], num2: array[i]))
}

可以使用var保留最后一个元素的引用,并使用defer在映射中更新它:

extension Collection where Element: FloatingPoint {
    var percentageDifference: [Element] {
        guard var last = first else { return [] }
        return map { element in
            defer { last = element }
            return (element - last) / last * 100
        }
    }
}

let numbers: [Double] = [4, 5, 2, 4, 3, 6]
let percentageDifference = numbers.percentageDifference
print(percentageDifference)   // "[0.0, 25.0, -60.0, 100.0, -25.0, 100.0]\n"

let array: [Double] = [4, 5, 2, 4, 3, 6]
let newArray = array.indices.map { i in (array[i] - array[max(0,i-1)]) / array[max(0, i-1)] * 100.0 }
print(newArray) // [0.0, 25.0, -60.0, 100.0, -25.0, 100.0]

有少量代码重复,但是它以一种性能良好的方式工作,我想说:)