是否有任何C#数学库进行插值/外推

时间:2010-03-07 13:11:52

标签: c# math

例如,我有分数

Y X
100 50
90 43
80 32

需要解决y = 50

Y X
1/1/2009 100
1/3/2009 97
1/4/2009 94
1/5/2009 92
1/6/2009 91
1/7/2009 89

需要解决y = 1/23/2009

4 个答案:

答案 0 :(得分:17)

我使用的是Math.NET http://numerics.mathdotnet.com/

的数字组件

它包含“各种插值方法,包括重心方法和样条”。

但俗话说,有谎言,该死的谎言和双三次样条插值。

答案 1 :(得分:2)

看看你是否能在ALGLIB找到你想要的东西。当然,您仍然需要为您的问题做出适当类型的插值/外推决策。

答案 2 :(得分:1)

我不知道库,但这里是一个简单的Secant求解器:

class SecantSolver
{
    private int     _maxSteps= 10;
    private double _precision= 0.1;

    public SecantSolver(int maxSteps, double precision)
    {
        _maxSteps= maxSteps;
        _precision= precision;

        if (maxSteps <= 0)
            throw new ArgumentException("maxSteps is out of range; must be greater than 0!");

        if (precision <= 0)
            throw new ArgumentException("precision is out of range; must be greater than 0!");

    }

    private double ComputeNextPoint(double p0, double p1, Func<Double,Double> f)
    {
        double r0 = f(p0);
        double r1 = f(p1);
        double p2 = p1 - r1 * (p1-p0) / (r1-r0); // the basic secant formula
        return p2;
    }

    public double Solve( double lowerBound, double upperBound, Func<Double,Double> f, out String message)
    {
        double p2,p1,p0;
        int i;
        p0=lowerBound;
        p1=upperBound;
        p2= ComputeNextPoint(p0,p1,f);

        // iterate till precision goal is met or the maximum
        // number of steps is reached
        for(i=0; System.Math.Abs(f(p2))>_precision &&i<_maxSteps;i++) {
            p0=p1;
            p1=p2;
            p2=ComputeNextPoint(p0,p1,f);
        }

        if (i < _maxSteps)
            message = String.Format("Method converges in " + i + " steps.");
        else
            message = String.Format("{0}. The method did not converge.", p2);

        return p2;
    }
}

用法:

SecantSolver solver= new SecantSolver(200,              // maxsteps
                                      0.00000001f/100   // tolerance
                                      );

string message;
double root= solver.Solve(0.10,   // initial guess (lower)
                          1.0,    // initial guess (upper)
                          f,      // the function to solve
                          out message
                          );

答案 3 :(得分:0)

ILNumerics Interpolation Toolbox带来所有标准插值函数。您将找到各种插补/外插功能,具有通用,简单的界面。与其他解决方案相比的一个特别优势是速度:这些功能经过精心优化,可以在多核硬件和大数据上尽可能快地进行优化。