Swift Accelerate for Mean&标准偏差

问题描述

我正在研究Accelerate以在Swift中计算数组的均值和标准差。

I am looking at Accelerate to compute mean and standard deviation of arrays in Swift.

我可以做均值。我如何做标准偏差？

I can do the mean. How do I do the standard deviation?

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

推荐答案

您可以计算平均值的标准差，以及
的均方值（比较和）：

You can compute the standard deviation from the mean value andthe mean square value (compare https://en.wikipedia.org/wiki/Standard_deviation#Identities_and_mathematical_properties and 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和更高版本或macOS 10.11和更高版本），使用：

Alternatively (for iOS 9.0 and later or macOS 10.11 and later), use 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

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