Question

我试图在一组点上使用多项式回归将它们映射到曲线。当我构建设计矩阵时，分配值似乎没有正确进行。构造的矩阵是3×n矩阵，其中n是点的数量，并且与设计矩阵一样，第一列应该全部为1（来自// HERE IS ASSIGNMENT部分）。当我打印矩阵时，第一列充满了随机数，通常是infs和nans。如何定期分配？这会导致使用矩阵进行任何数学运算时出现进一步的问题，因为它充满了数字和infs。

以下代码是构建矩阵的地方。

for (auto it = lane_lines.begin(); it != lane_lines.end(); it++) {
                        vector<pair<Point, Point> > lines = *it;

                        // Remove random line clusters
                        if (lines.size() < 10) {
                                it = --lane_lines.erase(it);
                                continue;
                        }   

                        Mat design(lines.size() * 2, 3, CV_64FC1);
                        Mat y_vec(lines.size() * 2, 1, CV_64FC1);

                        for (int i = 0; i < lines.size(); i += 2) {
                                // HERE IS ASSIGNMENT
                                design.at<double>(i * 2, 0) = 1.0;
                                design.at<double>(i * 2, 1) = lines[i].first.x;
                                design.at<double>(i * 2, 2) = pow(lines[i].first.x, 2); 
                                design.at<double>(i * 2 + 1, 0) = 1.0;
                                design.at<double>(i * 2 + 1, 1) = lines[i].second.x;
                                design.at<double>(i * 2 + 1, 2) = pow(lines[i].second.x, 2); 

                                y_vec.at<double>(i * 2, 0) = lines[i].first.y;
                                y_vec.at<double>(i * 2 + 1, 0) = lines[i].second.y;
                        }   

                        // TODO: Deal with all the NaNs
                        cout << design << endl;
                        cout << design.t() * design << endl;
                        cout << "DET: " << determinant(design.t() * design) << endl << endl;

                        Mat std_poly = ((design.t() * design).inv() * design.t()) * y_vec;

                        double a = std_poly.at<double>(0, 2); 
                        double b = std_poly.at<double>(0, 1); 
                        double c = std_poly.at<double>(0, 0); 
                        double h = -b / (2 * a == 0 ? numeric_limits<double>::min() : 2 * a); 
                        double k = c - (a * pow(h, 2));

                        tuple<double, double, double, Point> curve(a, b, c, Point(h, k));
                        curves.push_back(curve);
                }

编辑：对于不同的数据集，似乎随机出现问题。

Answer 1

首先看一下你的代码：

for (int i = 0; i < lines.size(); i += 2) {
     // HERE IS ASSIGNMENT
     design.at<double>(i * 2, 0) = 1.0;
     design.at<double>(i * 2, 1) = lines[i].first.x;
     design.at<double>(i * 2, 2) = pow(lines[i].first.x, 2); 
     design.at<double>(i * 2 + 1, 0) = 1.0;
     design.at<double>(i * 2 + 1, 1) = lines[i].second.x;
     design.at<double>(i * 2 + 1, 2) = pow(lines[i].second.x, 2); 
     y_vec.at<double>(i * 2, 0) = lines[i].first.y;
     y_vec.at<double>(i * 2 + 1, 0) = lines[i].second.y;
   }

对于第一次迭代i=0这两行

design.at<double>(i * 2, 0) = 1.0;
design.at<double>(i * 2 + 1, 0) = 1.0;

成为：

design.at<double>(0, 0) = 1.0;
design.at<double>(1, 0) = 1.0;

然后下一次迭代是i=2，因为循环中有i+=2所以上面的行变为：

design.at<double>(4, 0) = 1.0;
design.at<double>(5, 0) = 1.0;

我认为您应该将i+=2更改为i++以修复此行为。

NB 您的代码可能还有其他问题。

OpenCV中的矩阵值为inf / nan

1 个答案: