在F#中重写C#代码

Question

只是搞乱了F#,我试图创建一个基于这个C#版本的基本拉格朗日插值函数(从C++ wiki条目复制):

    double Lagrange(double[] pos, double[] val, double desiredPos)
    {
        double retVal = 0;

        for (int i = 0; i < val.Length; ++i)
        {
            double weight = 1;

            for (int j = 0; j < val.Length; ++j)
            {
                // The i-th term has to be skipped
                if (j != i)
                {
                    weight *= (desiredPos - pos[j]) / (pos[i] - pos[j]);
                }
            }

            retVal += weight * val[i];
        }

        return retVal;
    }

使用我对F#和函数式编程的有限知识,我能想到的最好的是:

let rec GetWeight desiredPos i j (pos : float[]) weight = 
   match i with
   | i when j = pos.Length -> weight
   | i when i = j -> GetWeight desiredPos i (j+1) pos weight 
   | i -> GetWeight desiredPos i (j+1) pos (weight * (desiredPos - pos.[j])/(pos.[i] - pos.[j]) ) 

let rec Lagrange (pos : float[]) (vals : float[]) desiredPos result counter = 
   match counter with
   | counter when counter = pos.Length -> result
   | counter -> Lagrange pos vals desiredPos (result + (GetWeight desiredPos counter 0 pos 1.0)* vals.[counter]) (counter+1)

有人可以根据相同的 C#代码提供更好/更整洁的F#版本吗？

Answer 1

折叠序列是用累加器替换循环的常见方法。

let Lagrange(pos:_[], v:_[], desiredPos) =
  seq {0 .. v.Length-1} 
  |> Seq.fold (fun retVal i -> 
      seq {for j in 0 .. pos.Length-1 do if i <> j then yield j} 
      |> Seq.fold (fun w j -> w * (desiredPos - pos.[j]) / (pos.[i] - pos.[j])) 1.0
      |> (fun weight -> weight * v.[i] + retVal)) 0.0