我意识到,当我使用“la_solve”时,由于“la_solve”可以处理的样本数量有限,我会收到 Malloc 错误。
我注意到如果我使用大约 900 个样本,程序运行良好。
因此,我修改了我的代码,以在样本 > 901 的情况下查看 901 个样本的连续 block 中的整个数据。
但是,我使用了相当幼稚的技术,并且确信这可以变得更加高效和稳健:(1) 我需要一个条件,如果样本小于 900,则不要分块; (2) 在附加分块数据时注意 x 和 y 值的重复; (3) 使算法更加高效、快速和稳健。
我最需要帮助的是代码末尾的“//调用重新采样器功能”部分
谢谢!!!
import UIKit
import Accelerate.vecLib.LinearAlgebra
public class Resample {
public class func resample(xi: [Double], xii: [Double], a: [Double]) ->[Double]
{
// xi - original time stamps (original x)
// xii - desired time stamps (desired x)
// a - orignal y values
// ->[Double] - interpolated y values
let ni = xii.count // Desired x count
let n = xi.count // Actual x count
var h: [Double] = Array(repeating:0, count:n-1)
for j in 0..<n-1 {
h[j] = (xi[j+1] - xi[j])
}
var A: [Double] = Array(repeating:0, count:(n)*(n))
A[0] = 1.0
A[(n*n)-1] = 1.0
for i in 1..<(n-1) {
A[(n+1)*i-1] = h[i-1]
A[(n+1)*i] = 2*(h[i-1] + h[i])
A[(n+1)*i+1] = h[i]
}
var b: [Double] = Array(repeating:0, count:n)
for i in 1..<n-1 {
b[i] = (3/h[i])*(a[i+1]-a[i]) - (3/h[i-1])*(a[i]-a[i-1])
}
let matA = la_matrix_from_double_buffer(A, la_count_t(n), la_count_t(n), la_count_t(n), la_hint_t(LA_NO_HINT), la_attribute_t(LA_DEFAULT_ATTRIBUTES))
let vecB = la_matrix_from_double_buffer(b, la_count_t(n), 1, 1, la_hint_t(LA_NO_HINT), la_attribute_t(LA_DEFAULT_ATTRIBUTES))
let vecCj = la_solve(matA, vecB)
var cj: [Double] = Array(repeating: 0.0, count: n)
let status = la_matrix_to_double_buffer(&cj, 1, vecCj)
if status == la_status_t(LA_SUCCESS) {
//print(cj.count)
}
else {
print("Failure: \(status)")
}
var bj: [Double] = Array(repeating:0, count:n-1)
for i in 0..<n-1 {
bj[i] = (1/h[i])*(a[i+1]-a[i]) - (1/3*h[i])*(2*cj[i]+cj[i+1])
}
var dj: [Double] = Array(repeating:0, count:n-1)
for i in 0..<n-1 {
dj[i] = (1/(3*h[i])) * (cj[i+1]-cj[i])
}
var P: [Double] = Array(repeating: 0.0, count: (n-1)*4)
for i in 0..<n-1 {
P[(i*4)] = dj[i]
P[(i*4)+1] = cj[i]
P[(i*4)+2] = bj[i]
P[(i*4)+3] = a[i]
}
var ai: [Double] = Array(repeating: 0.0, count: ni)
var jl = Double(1)
for i in 0..<ni {
let inter = Double(xii[i])
while (inter > xi[Int(jl)] && (Int(jl) < n)) {
jl += 1
}
let ind = 4*Int(jl)-4
var pp = Array(P[ind...ind+3])
let xx = Double(xii[i]-xi[Int(jl)-1])
ai[i] = pp[0]*pow(xx, 3) + pp[1]*pow(xx, 2) + pp[2]*xx + pp[3]
}
return(ai)
}
}
// Calling the Re-sampler FUNCTION
let x = Array(stride(from: 0, through: 16649, by: 1.0)) // Orig X (Given, say)
let y = Array(stride(from: 0, through: 16649, by: 1.0)) // Orig Y (Given, say)
let f = 2.0 // Hz (Desired Sampling Freq., say)
let g = Array(stride(from: 0, through: x.count-1, by: 900))
// Helper array for chunking (901 samples each time, say)
var xxx = [Double]() // Results for appended chunks X
var yyy = [Double]() // Results for appended chunks Y
for i in 0..<g.count-1 { // loop through
let xc = Array(x[Int(g[i])...Int(g[i+1])]) // x chunk (count = 901)
let yc = Array(y[Int(g[i])...Int(g[i+1])]) // y chunk (count = 901)
let xci = Array(stride(from: xc[0], through: xc[xc.count-1], by: 1.0/f))
// Desired time stamps for chunk of 901 samples
let yci = Resample.resample(xi: xc, xii: xci, a: yc) // Interpolate chunk
xxx += xci // Append interpolation X
yyy += yci // Append interpolation X
}
if(Int(g[g.count-1])<x.count){
// If helper function last index < original no. of samples
let glb = (Int(g[g.count-1])) // last chunk begin index
let gle = (x.count) // last chunk end index
let xce = Array(x[glb...gle-1]) // last x chunk
let yce = Array(y[glb...gle-1]) // last y chunk
let xcei = Array(stride(from: xce[0], through: xce[xce.count-1], by: 1.0/f))
// Desired time stamps for last chunk of samples
let ycei = Resample.resample(xi: xce, xii: xcei, a: yce) // Interpolate last chunk
xxx += xcei // Append interpolation X for last chunk
yyy += ycei // Append interpolation X for last chunk
}
print(xxx) // would like to identify repeated x values and remove them
print(yyy) // remove y values corresponsding to repated x values (as found above)
// Calling the Re-sampler FUNCTION
最佳答案
请注意,您正在使用二进制搜索寻找每个 ind
,而它的值是单调递增的。所以你可以只检查下一个 x 值是否保持在相同的间隔内或移动到下一个。伪代码:
jl = 1 //before cycle
//inside cycle
while (current_x > xi[jl]) and (jl < n)
jl++
ind = 4 * jl - 4
la_solve
是求解线性方程组的通用方法吗?如果是,那么您最好对三对角线系统使用专用方法(三次复杂度与线性复杂度)。 Classic text . Python implementation . Wiki .对于长输入数组很重要。
关于swift - 如何使三次样条插值算法更快?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/41542256/