我在插入数据文件时出现问题，我已将其从.csv转换为X数组和Y数组，其中X [0]对应于点Y [0]。
我需要在值之间进行插值，以便在最后给出一个平滑的文件。 我正在使用Picoscope输出函数，这意味着每一行的时间间隔相等，所以只使用Y值，这就是为什么我在你看到我的代码时会以一种奇怪的方式尝试这样做。
它必须处理的价值有：
X Y
0 0
2.5 0
2.5 12000
7.5 12000
7.5 3000
10 3000
10 6000
11 6625.254
12 7095.154
所以，彼此相邻的2个Y值是相同的，它们之间是一条直线，但是当它们像病房的X = 10那样变化时，在这个例子中变成正弦波。
我一直在寻找插值等方程式以及此处的其他帖子，但我多年没有做过那种数学，遗憾的是我再也无法弄清楚了，因为有2个未知数我可以'想想如何将其编入c＃。
我拥有的是：
int a = 0, d = 0, q = 0;
bool up = false;
double times = 0, change = 0, points = 0, pointchange = 0;
double[] newy = new double[8192];
while (a < sizeoffile-1 && d < 8192)
{
Console.Write("...");
if (Y[a] == Y[a+1])//if the 2 values are the same add correct number of lines in to make it last the correct time
{
times = (X[a] - X[a + 1]);//number of repetitions
if ((X[a + 1] - X[a]) > times)//if that was a negative number try the other way around
times = (X[a + 1] - X[a]);//number of repetitions
do
{
newy[d] = Y[a];//add the values to the newy array which replaces y later on in the program
d++;//increment newy position
q++;//reduce number of reps in this loop
}
while (q < times + 1 && d < 8192);
q = 0;//reset reps
}
else if (Y[a] != Y[a + 1])//if the 2 values are not the same interpolate between them
{
change = (Y[a] - Y[a + 1]);//work out difference between the values
up = true;//the waveform is increasing
if ((Y[a + 1] - Y[a]) > change)//if that number was a negative try the other way around
{
change = (Y[a + 1] - Y[a]);//work out difference bwteen values
up = false;//the waveform is decreasing
}
points = (X[a] - X[a + 1]);//work out amount of time between given points
if ((X[a + 1] - X[a]) > points)//if that was a negative try other way around
points = (X[a + 1] - X[a]);///work out amount of time between given points
pointchange = (change / points);//calculate the amount per point in time the value changes by
if (points > 1)//any lower and the values cause errors
{
newy[d] = Y[a];//load first point
d++;
do
{
if (up == true)//
newy[d] = ((newy[d - 1]) + pointchange);
else if (up == false)
newy[d] = ((newy[d - 1]) - pointchange);
d++;
q++;
}
while (q < points + 1 && d < 8192);
q = 0;
}
else if (points != 0 && points > 0)
{
newy[d] = Y[a];//load first point
d++;
}
}
a++;
}
这会产生一个接近的波形，但它仍然非常陡峭。
所以有人能看出为什么这不是很准确吗？ 如何提高其准确性？ 或者使用数组的另一种方法呢？
感谢您寻找：）
答案 0 :(得分：11)
为我试试这个方法：
static public double linear(double x, double x0, double x1, double y0, double y1)
{
if ((x1 - x0) == 0)
{
return (y0 + y1) / 2;
}
return y0 + (x - x0) * (y1 - y0) / (x1 - x0);
}
实际上你应该能够拿出你的阵列并像这样使用它：
var newY = linear(X[0], X[0], X[1], Y[0], Y[1]);
我从here中提取了代码，但确认该算法符合理论here，因此我 认为 是正确的。但是，你可能应该考虑使用多项式插值，如果它仍然很稳定，请注意理论链接，它表明线性插值产生了陡峭的波。
所以，我给出的第一个链接，我从中获取了这个代码，也有一个多项式算法：
static public double lagrange(double x, double[] xd, double[] yd)
{
if (xd.Length != yd.Length)
{
throw new ArgumentException("Arrays must be of equal length."); //$NON-NLS-1$
}
double sum = 0;
for (int i = 0, n = xd.Length; i < n; i++)
{
if (x - xd[i] == 0)
{
return yd[i];
}
double product = yd[i];
for (int j = 0; j < n; j++)
{
if ((i == j) || (xd[i] - xd[j] == 0))
{
continue;
}
product *= (x - xd[i]) / (xd[i] - xd[j]);
}
sum += product;
}
return sum;
}
要使用这个，你将不得不决定如何提升你的x值，所以我们想通过找到当前迭代和下一个迭代之间的中点来做到这一点： / p>
for (int i = 0; i < X.Length; i++)
{
var newY = lagrange(new double[] { X[i]d, X[i+1]d }.Average(), X, Y);
}
请注意，此循环会有更多内容，例如确保有i+1等等，但我想知道是否可以给你一个开始。
答案 1 :(得分：1)
下面的解决方案计算具有相同X的给定点的Y值的平均值，就像Matlab polyfit函数一样
Linq和.NET框架版本＆gt;这个swift API必须使用3.5。 代码中的注释。
using System;
using System.Collections.Generic;
using System.Linq;
/// <summary>
/// Linear Interpolation using the least squares method
/// <remarks>http://mathworld.wolfram.com/LeastSquaresFitting.html</remarks>
/// </summary>
public class LinearLeastSquaresInterpolation
{
/// <summary>
/// point list constructor
/// </summary>
/// <param name="points">points list</param>
public LinearLeastSquaresInterpolation(IEnumerable<Point> points)
{
Points = points;
}
/// <summary>
/// abscissae/ordinates constructor
/// </summary>
/// <param name="x">abscissae</param>
/// <param name="y">ordinates</param>
public LinearLeastSquaresInterpolation(IEnumerable<float> x, IEnumerable<float> y)
{
if (x.Empty() || y.Empty())
throw new ArgumentNullException("null-x");
if (y.Empty())
throw new ArgumentNullException("null-y");
if (x.Count() != y.Count())
throw new ArgumentException("diff-count");
Points = x.Zip(y, (unx, uny) => new Point { x = unx, y = uny } );
}
private IEnumerable<Point> Points;
/// <summary>
/// original points count
/// </summary>
public int Count { get { return Points.Count(); } }
/// <summary>
/// group points with equal x value, average group y value
/// </summary>
public IEnumerable<Point> UniquePoints
{
get
{
var grp = Points.GroupBy((p) => { return p.x; });
foreach (IGrouping<float,Point> g in grp)
{
float currentX = g.Key;
float averageYforX = g.Select(p => p.y).Average();
yield return new Point() { x = currentX, y = averageYforX };
}
}
}
/// <summary>
/// count of point set used for interpolation
/// </summary>
public int CountUnique { get { return UniquePoints.Count(); } }
/// <summary>
/// abscissae
/// </summary>
public IEnumerable<float> X { get { return UniquePoints.Select(p => p.x); } }
/// <summary>
/// ordinates
/// </summary>
public IEnumerable<float> Y { get { return UniquePoints.Select(p => p.y); } }
/// <summary>
/// x mean
/// </summary>
public float AverageX { get { return X.Average(); } }
/// <summary>
/// y mean
/// </summary>
public float AverageY { get { return Y.Average(); } }
/// <summary>
/// the computed slope, aka regression coefficient
/// </summary>
public float Slope { get { return ssxy / ssxx; } }
// dotvector(x,y)-n*avgx*avgy
float ssxy { get { return X.DotProduct(Y) - CountUnique * AverageX * AverageY; } }
//sum squares x - n * square avgx
float ssxx { get { return X.DotProduct(X) - CountUnique * AverageX * AverageX; } }
/// <summary>
/// computed intercept
/// </summary>
public float Intercept { get { return AverageY - Slope * AverageX; } }
public override string ToString()
{
return string.Format("slope:{0:F02} intercept:{1:F02}", Slope, Intercept);
}
}
/// <summary>
/// any given point
/// </summary>
public class Point
{
public float x { get; set; }
public float y { get; set; }
}
/// <summary>
/// Linq extensions
/// </summary>
public static class Extensions
{
/// <summary>
/// dot vector product
/// </summary>
/// <param name="a">input</param>
/// <param name="b">input</param>
/// <returns>dot product of 2 inputs</returns>
public static float DotProduct(this IEnumerable<float> a, IEnumerable<float> b)
{
return a.Zip(b, (d1, d2) => d1 * d2).Sum();
}
/// <summary>
/// is empty enumerable
/// </summary>
/// <typeparam name="T"></typeparam>
/// <param name="a"></param>
/// <returns></returns>
public static bool Empty<T>(this IEnumerable<T> a)
{
return a == null || a.Count() == 0;
}
}
使用它：
var llsi = new LinearLeastSquaresInterpolation(new Point[]
{
new Point {x=1, y=1}, new Point {x=1, y=1.1f}, new Point {x=1, y=0.9f},
new Point {x=2, y=2}, new Point {x=2, y=2.1f}, new Point {x=2, y=1.9f},
new Point {x=3, y=3}, new Point {x=3, y=3.1f}, new Point {x=3, y=2.9f},
new Point {x=10, y=10}, new Point {x=10, y=10.1f}, new Point {x=10, y=9.9f},
new Point {x=100, y=100}, new Point{x=100, y=100.1f}, new Point {x=100, y=99.9f}
});
或者：
var llsi = new LinearLeastSquaresInterpolation(
new float[] { 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9 },
new float[] { 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9 });