Question

我正在考虑使用Accelerate来计算Swift中数组的平均值和标准偏差。

我可以做到这一点。我该如何做标准偏差？

let rr: [Double] = [ 18.0, 21.0, 41.0, 42.0, 48.0, 50.0, 55.0, 90.0 ]

var mn: Double = 0.0

vDSP_meanvD(rr, 1, &mn, vDSP_Length(rr.count))

print(mn) // prints correct mean as 45.6250

// Standard Deviation should be 22.3155

Answer 1

您可以计算平均值的标准偏差均方值（比较https://en.wikipedia.org/wiki/Standard_deviation#Identities_and_mathematical_properties和https://en.wikipedia.org/wiki/Algebraic_formula_for_the_variance）：

import Accelerate

let rr: [Double] = [ 18.0, 21.0, 41.0, 42.0, 48.0, 50.0, 55.0, 90.0 ]

var mn: Double = 0.0 // mean value
vDSP_meanvD(rr, 1, &mn, vDSP_Length(rr.count))

var ms: Double = 0.0 // mean square value
vDSP_measqvD(rr, 1, &ms, vDSP_Length(rr.count))

let sddev = sqrt(ms - mn * mn) * sqrt(Double(rr.count)/Double(rr.count - 1))

print(mn, sddev)
// 45.625 22.315513501982

或者（对于iOS 9.0及更高版本或macOS 10.11及更高版本），请使用vDSP_normalizeD：

var mn = 0.0
var sddev = 0.0
vDSP_normalizeD(rr, 1, nil, 1, &mn, &sddev, vDSP_Length(rr.count))
sddev *= sqrt(Double(rr.count)/Double(rr.count - 1))

print(mn, sddev)
// 45.625 22.315513501982

Answer 2

@Martin R答案的附加组件：还有一个用于浮点/单精度的 vDSP_normalize 函数。

func vDSP_normalize(UnsafePointer<Float>, vDSP_Stride, UnsafeMutablePointer<Float>?, vDSP_Stride, UnsafeMutablePointer<Float>, UnsafeMutablePointer<Float>, vDSP_Length)
//Compute mean and standard deviation and then calculate new elements to have a zero mean and a unit standard deviation. Single precision.

func vDSP_normalizeD(UnsafePointer<Double>, vDSP_Stride, UnsafeMutablePointer<Double>?, vDSP_Stride, UnsafeMutablePointer<Double>, UnsafeMutablePointer<Double>, vDSP_Length)
//Compute mean and standard deviation and then calculate new elements to have a zero mean and a unit standard deviation. Double precision.

Swift Accelerate for Mean＆amp;标准偏差

2 个答案: