c++ - 矩阵值在OpenCV中为inf/nan

我正在尝试对一组点使用多项式回归以将它们映射到曲线。当我构造设计矩阵时，分配值似乎没有正确处理。构造的矩阵是3 x n矩阵，其中n是点数，并且与设计矩阵一样，第一列应为全1(来自// HERE IS ASSIGNMENT部分)。当我打印矩阵时，第一列充满随机数，经常是infs和nans。常量分配怎么可能？由于矩阵中没有数字和infs，因此对矩阵进行数学运算会导致其他问题。

下面的代码是构造矩阵的位置。

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:问题似乎是针对不同的数据集随机出现的。

最佳答案

首先看一下您的代码:

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;

接下来的迭代是i=2，因为i+=2处于循环中，因此上述几行变为:

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++以解决此问题。

注意:您的代码可能还有其他问题。

关于c++ - 矩阵值在OpenCV中为inf/nan，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/49692515/