日韩性视频-久久久蜜桃-www中文字幕-在线中文字幕av-亚洲欧美一区二区三区四区-撸久久-香蕉视频一区-久久无码精品丰满人妻-国产高潮av-激情福利社-日韩av网址大全-国产精品久久999-日本五十路在线-性欧美在线-久久99精品波多结衣一区-男女午夜免费视频-黑人极品ⅴideos精品欧美棵-人人妻人人澡人人爽精品欧美一区-日韩一区在线看-欧美a级在线免费观看

歡迎訪問 生活随笔!

生活随笔

當前位置: 首頁 > 编程资源 > 编程问答 >内容正文

编程问答

拉格朗日插值拟合

發(fā)布時間:2023/12/15 编程问答 28 豆豆
生活随笔 收集整理的這篇文章主要介紹了 拉格朗日插值拟合 小編覺得挺不錯的,現(xiàn)在分享給大家,幫大家做個參考.



pre define

typedef struct Operators {std::vector<float> Operatos;}Operators;// 定義一個點 class PointF { public:PointF(){X = Y = 0.0f;}PointF(IN const PointF &point){X = point.X;Y = point.Y;}PointF(IN float x,IN float y){X = x;Y = y;}PointF operator+(IN const PointF& point) const{return PointF(X + point.X,Y + point.Y);}PointF operator-(IN const PointF& point) const{return PointF(X - point.X,Y - point.Y);}BOOL Equals(IN const PointF& point){return (X == point.X) && (Y == point.Y);}public:float X;float Y; };

fit lagrange

void CWinaTVWaveformDlg::fitLagrange(std::vector<PointF> pointList) {std::vector<Operators> OpList;// result std::vector<float> Xs;std::vector<float> Ys;try{if (pointList.size() > 0){//compute lagrange operator for each X coordinatefor (int x = 1; x < 2000; x++){//list of float to hold the Lagrange operators, Init the list with 1'sstd::vector<float> L(pointList.size(), 1); for (int i = 0; i < L.size(); i++){for (int k = 0; k < pointList.size(); k++){if (i != k)L[i] *= (float)(x - pointList[k].X) / (pointList[i].X - pointList[k].X);}}Operators o;o.Operatos = L;OpList.push_back(o);Xs.push_back(x);}//Computing the Polynomial P(x) which is y in our curvestd::vector<Operators>::iterator iter;for ( iter = OpList.begin(); iter != OpList.end(); iter++){float y = 0;for (int i = 0; i < pointList.size(); i++){y += iter->Operatos[i] * pointList[i].Y;}Ys.push_back(y);}//Drawing the curve in the simplest way}else{AfxMessageBox(_T("Lagrange curve fitting, add some points"));}}catch (CException* e){return;} }



總結(jié)

以上是生活随笔為你收集整理的拉格朗日插值拟合的全部內(nèi)容,希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯,歡迎將生活随笔推薦給好友。