swift - 以波动的时间间隔捕获的线性重采样数据点，到固定的时间间隔， swift

Question

我想将一些在波动的时间捕获的指标线性插值到固定的时间间隔。

let original_times:[Double] = [0.0,1.3,2.2,3.4,4.2,5.5,6.6,7.2,8.4,9.5,10.0]
let metric_1:[Double] = [4,3,6,7,4,5,7,4,2,7,2]

let wanted_times:[Double] = [0,1,2,3,4,5,6,7,8,9,10]

//linearly resample metric_1 (with corresponding sampling times 'original_times') to fixed time interval times 'wanted_times'

Accelerate 优惠 vDSP_vlint但我正在努力弄清楚如何为我的应用程序实现它。

func vDSP_vlint(_ __A: UnsafePointer<Float>, _ __B: UnsafePointer<Float>, _ __IB: vDSP_Stride, _ __C: UnsafeMutablePointer<Float>, _ __IC: vDSP_Stride, _ __N: vDSP_Length, _ __M: vDSP_Length)

Answer 1

我不是 100% 理解您想做的数学运算，但我确实理解如何使用 Accelerate。我创建了一个函数，可以更轻松地调用此 Accelerate 函数并向您展示它是如何工作的。

/**
 Vector linear interpolation between neighboring elements

 - Parameter a: Input vector.
 - Parameter b: Input vector: integer parts are indices into a and fractional parts are interpolation constants.

 Performs the following operation:

 ```C
 for (n = 0; n < N; ++n) {
    double b = B[n];
    double index = trunc([b]); //int part of B value
    double alpha = b - index; //frac part of B value

    double a0 = A[(int)index];     //indexed A value
    double a1 = A[(int)index + 1]; //next A value

    C[n] = a0 + (alpha * (a1 -a0)); //interpolated value
 }
 ```
 Generates vector C by interpolating between neighboring values of vector A as controlled by vector B. The integer portion of each element in B is the zero-based index of the first element of a pair of adjacent values in vector A.

 The value of the corresponding element of C is derived from these two values by linear interpolation, using the fractional part of the value in B.
*/
func interpolate(inout a: [Double], inout b: [Double]) -> [Double] {
    var c = [Double](count: b.count, repeatedValue: 0)
    vDSP_vlintD(&a, &b, 1, &c, 1, UInt(b.count), UInt(a.count))
    return c
}

编辑:好的，我全神贯注于你的问题，我现在明白你想做什么了。这样做很有趣，我想到了这个:

import Accelerate

func calculateB(sampleTimes: [Double], outputTimes: [Double]) -> [Double] {
    var i = 0
    return outputTimes.map { (time: Double) -> Double in
        defer {
            if time > sampleTimes[i] { i++ }
        }
        return Double(i) + (time - sampleTimes[i]) / (sampleTimes[i+1] - sampleTimes[i])
    }
}

func interpolate(inout b: [Double], inout data: [Double]) -> [Double] {
    var c = [Double](count: b.count, repeatedValue: 0)
    vDSP_vlintD(&data, &b, 1, &c, 1, UInt(b.count), UInt(data.count))
    return c
}


let sampleTimes : [Double] = [0.0, 1.3, 2.2, 3.4, 4.2, 5.5, 6.6, 7.2, 8.4, 9.5, 10.0]
let outputTimes : [Double] = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

var metric_1 : [Double] = [4, 3, 6, 7, 4, 5, 7, 4, 2, 7, 2]
var metric_2 : [Double] = [5, 4, 7, 5, 6, 6, 1, 3, 1, 6, 7]
var metric_3 : [Double] = [9, 8, 5, 7, 4, 8, 5, 6, 8, 9, 5]

var b = calculateB(sampleTimes, outputTimes: outputTimes)

interpolate(&b, data: &metric_1)   // [4, 3.230769, 5.333333, 6.666667, 4.75, 4.615385, 5.909091, 5, 2.666667, 4.727273, 2]
interpolate(&b, data: &metric_2)   // [5, 4.230769, 6.333333, 5.666667, 5.75, 6, 3.727273, 2.333333, 1.666667, 3.727273, 7]
interpolate(&b, data: &metric_3)   // [9, 8.230769, 5.666667, 6.333333, 4.75, 6.461538, 6.636364, 5.666667, 7.333333, 8.545455, 5]

这些变量是 Accelerate 所必需的。我不知道 calculateB 如何用 Accelerate 完成，我的意思是我认为这是可能的，但是搜索正确的 vDSP 函数是一件痛苦的事情......