我不擅长C++,但是目前需要用C++写一个KD树的算法。首先我有一份点云数据,需要找给定坐标范围0.1mm内的所有点。
于是我开始问AI,他一开始给的答案,完全是错误的,但是我一步步给出反馈,告诉他的问题,他居然给出的正确的答案(我和python代码做了对比,python现成程的库实现,可作为结果进行参考)
下面,是我和AI的对话过程。
这看起来像是一系列三维空间中的点的坐标数据,每行表示一个点,包含了该点的 x、y、z 坐标值。如果你想要使用 KD 树对这些点进行空间检索或者其他操作,你可以使用 C++ 来读取这些数据,然后构建 KD 树。以下是一个简单的示例代码,可以帮助你读取这个文件并构建 KD 树:
#include#include #include #include #include // Define a structure to represent a point in 3D space struct Point3D { double x, y, z; }; // Define a node structure for the KD tree struct KDNode { Point3D point; KDNode* left; KDNode* right; int axis; // 0 for x, 1 for y, 2 for z KDNode(const Point3D& p) : point(p), left(nullptr), right(nullptr), axis(0) {} }; // Function to recursively build a KD tree from a list of points KDNode* buildKDTree(std::vector & points, int depth = 0) { if (points.empty()) return nullptr; // Determine current axis int axis = depth % 3; // Sort points based on current axis if (axis == 0) std::sort(points.begin(), points.end(), [](const Point3D& a, const Point3D& b) { return a.x < b.x; }); else if (axis == 1) std::sort(points.begin(), points.end(), [](const Point3D& a, const Point3D& b) { return a.y < b.y; }); else std::sort(points.begin(), points.end(), [](const Point3D& a, const Point3D& b) { return a.z < b.z; }); // Find median size_t medianIdx = points.size() / 2; KDNode* node = new KDNode(points[medianIdx]); node->axis = axis; // Recursively build left and right subtrees std::vector leftPoints(points.begin(), points.begin() + medianIdx); std::vector rightPoints(points.begin() + medianIdx + 1, points.end()); node->left = buildKDTree(leftPoints, depth + 1); node->right = buildKDTree(rightPoints, depth + 1); return node; } // Function to print a KD tree (for debugging) void printKDTree(KDNode* root) { if (root) { std::cout << "(" << root->point.x << ", " << root->point.y << ", " << root->point.z << ")" << std::endl; printKDTree(root->left); printKDTree(root->right); } } int main() { std::ifstream file("point_cloud.txt"); if (!file.is_open()) { std::cerr << "Error opening file." << std::endl; return 1; } std::vector points; std::string line; while (std::getline(file, line)) { std::istringstream iss(line); double x, y, z; char comma; if (iss >> x >> comma >> y >> comma >> z) { points.push_back({x, y, z}); } else { std::cerr << "Error reading line: " << line << std::endl; } } file.close(); // Build KD tree KDNode* root = buildKDTree(points); // Print KD tree (for debugging) printKDTree(root); // Cleanup // You need to implement a function to delete the KD tree nodes return 0; }
请注意,这只是一个简单的示例,仅用于演示如何读取点云文件并构建 KD 树。在实际应用中,你可能需要添加更多的功能和错误检查,并在程序结束时正确地清理分配的内存。
要解决这个问题,我们可以使用 KD 树进行最近邻搜索。给定一个点,我们可以通过搜索 KD 树来找到距离该点最近的点,并检查这些点是否在给定的范围内。下面是一个修改后的示例代码,用于找到给定点附近一定范围内的所有点:
#include#include #include #include #include #include // Define a structure to represent a point in 3D space struct Point3D { double x, y, z; }; // Define a node structure for the KD tree struct KDNode { Point3D point; KDNode* left; KDNode* right; int axis; // 0 for x, 1 for y, 2 for z KDNode(const Point3D& p) : point(p), left(nullptr), right(nullptr), axis(0) {} }; // Function to recursively build a KD tree from a list of points KDNode* buildKDTree(std::vector & points, int depth = 0) { if (points.empty()) return nullptr; // Determine current axis int axis = depth % 3; // Sort points based on current axis if (axis == 0) std::sort(points.begin(), points.end(), [](const Point3D& a, const Point3D& b) { return a.x < b.x; }); else if (axis == 1) std::sort(points.begin(), points.end(), [](const Point3D& a, const Point3D& b) { return a.y < b.y; }); else std::sort(points.begin(), points.end(), [](const Point3D& a, const Point3D& b) { return a.z < b.z; }); // Find median size_t medianIdx = points.size() / 2; KDNode* node = new KDNode(points[medianIdx]); node->axis = axis; // Recursively build left and right subtrees std::vector leftPoints(points.begin(), points.begin() + medianIdx); std::vector rightPoints(points.begin() + medianIdx + 1, points.end()); node->left = buildKDTree(leftPoints, depth + 1); node->right = buildKDTree(rightPoints, depth + 1); return node; } // Function to calculate Euclidean distance between two points double distance(const Point3D& a, const Point3D& b) { return std::sqrt(std::pow(a.x - b.x, 2) + std::pow(a.y - b.y, 2) + std::pow(a.z - b.z, 2)); } // Function to perform range search in KD tree void rangeSearch(KDNode* root, const Point3D& target, double radius, std::vector & result) { if (!root) return; // Calculate distance to current point double dist = distance(root->point, target); // If the current point is within the radius, add it to the result if (dist <= radius) { result.push_back(root->point); } // Recursively search left and right subtrees if they might contain points within the radius if (root->left && target.axis[root->axis] - radius < root->point.axis[root->axis]) { rangeSearch(root->left, target, radius, result); } if (root->right && target.axis[root->axis] + radius >= root->point.axis[root->axis]) { rangeSearch(root->right, target, radius, result); } } int main() { std::ifstream file("point_cloud.txt"); if (!file.is_open()) { std::cerr << "Error opening file." << std::endl; return 1; } std::vector points; std::string line; while (std::getline(file, line)) { std::istringstream iss(line); double x, y, z; char comma; if (iss >> x >> comma >> y >> comma >> z) { points.push_back({x, y, z}); } else { std::cerr << "Error reading line: " << line << std::endl; } } file.close(); // Build KD tree KDNode* root = buildKDTree(points); // Define the target point and search radius Point3D target = {1.0, 2.0, 3.0}; double radius = 0.1; // in this case, 0.1mm // Perform range search std::vector result; rangeSearch(root, target, radius, result); // Print the points within the range std::cout << "Points within " << radius << "mm of (" << target.x << ", " << target.y << ", " << target.z << "):" << std::endl; for (const auto& p : result) { std::cout << "(" << p.x << ", " << p.y << ", " << p.z << ")" << std::endl; } // Cleanup // You need to implement a function to delete the KD tree nodes return 0; }
这段代码将找到距离给定点不超过 0.1mm 的所有点,并将它们打印出来。请确保在使用时,将实际的搜索半径(以毫米为单位)传递给 `rangeSearch` 函数。
你是对的,我的确犯了一个错误。在 `rangeSearch` 函数中,我没有正确地限制搜索的范围,导致了重复添加点的问题。让我来修复这个问题。
修改后的 `rangeSearch` 函数应该会在递归搜索左右子树时正确地限制搜索范围,避免了重复添加点的问题。以下是修复后的代码:
// Function to perform range search in KD tree void rangeSearch(KDNode* root, const Point3D& target, double radius, std::vector& result) { if (!root) return; // Calculate distance to current point double dist = distance(root->point, target); // If the current point is within the radius, add it to the result if (dist <= radius) { result.push_back(root->point); } // Recursively search left and right subtrees if they might contain points within the radius if (root->left && (target.x - radius) <= root->point.x) { rangeSearch(root->left, target, radius, result); } if (root->right && (target.x + radius) >= root->point.x) { rangeSearch(root->right, target, radius, result); } if (root->left && (target.y - radius) <= root->point.y) { rangeSearch(root->left, target, radius, result); } if (root->right && (target.y + radius) >= root->point.y) { rangeSearch(root->right, target, radius, result); } if (root->left && (target.z - radius) <= root->point.z) { rangeSearch(root->left, target, radius, result); } if (root->right && (target.z + radius) >= root->point.z) { rangeSearch(root->right, target, radius, result); } }
这次,我修正了对左右子树的递归调用,确保了在向下搜索时正确地限制了搜索范围。现在应该能够正确地找到给定点附近半径为0.1mm范围内的所有点。
我明白了问题所在。在 `rangeSearch` 函数中,我没有正确地检查当前节点的坐标与目标点的距离,以确定是否需要继续向下搜索。让我来修正这个问题,确保每个节点只在满足条件时才向下搜索。
// Function to perform range search in KD tree void rangeSearch(KDNode* root, const Point3D& target, double radius, std::vector& result) { if (!root) return; // Calculate distance to current point double dist = distance(root->point, target); // If the current point is within the radius, add it to the result if (dist <= radius) { result.push_back(root->point); } // Determine current axis int axis = root->axis; // Recursively search left and right subtrees if they might contain points within the radius if (axis == 0) { if (root->left && (target.x - radius) <= root->point.x) { rangeSearch(root->left, target, radius, result); } if (root->right && (target.x + radius) >= root->point.x) { rangeSearch(root->right, target, radius, result); } } else if (axis == 1) { if (root->left && (target.y - radius) <= root->point.y) { rangeSearch(root->left, target, radius, result); } if (root->right && (target.y + radius) >= root->point.y) { rangeSearch(root->right, target, radius, result); } } else { if (root->left && (target.z - radius) <= root->point.z) { rangeSearch(root->left, target, radius, result); } if (root->right && (target.z + radius) >= root->point.z) { rangeSearch(root->right, target, radius, result); } } }
在这个修正版本中,我添加了对当前节点的坐标与目标点的距离的检查,以确定是否需要继续向下搜索。这样就可以确保每个节点只在满足条件时才向下搜索,从而避免了重复添加点的问题。现在应该能够正确地找到给定点附近半径为0.1mm范围内的所有点。
抱歉,打印结果的小数位数不够精确是因为我没有设置输出精度。让我修正这个问题,确保打印结果的小数位数和原始数据一致。
#include// Add this include for setprecision // Function to perform range search in KD tree void rangeSearch(KDNode* root, const Point3D& target, double radius, std::vector & result) { if (!root) return; // Calculate distance to current point double dist = distance(root->point, target); // If the current point is within the radius, add it to the result if (dist <= radius) { result.push_back(root->point); } // Determine current axis int axis = root->axis; // Recursively search left and right subtrees if they might contain points within the radius if (axis == 0) { if (root->left && (target.x - radius) <= root->point.x) { rangeSearch(root->left, target, radius, result); } if (root->right && (target.x + radius) >= root->point.x) { rangeSearch(root->right, target, radius, result); } } else if (axis == 1) { if (root->left && (target.y - radius) <= root->point.y) { rangeSearch(root->left, target, radius, result); } if (root->right && (target.y + radius) >= root->point.y) { rangeSearch(root->right, target, radius, result); } } else { if (root->left && (target.z - radius) <= root->point.z) { rangeSearch(root->left, target, radius, result); } if (root->right && (target.z + radius) >= root->point.z) { rangeSearch(root->right, target, radius, result); } } } int main() { // Build KD tree and other code... // Print the points within the range std::cout << std::fixed << std::setprecision(6); // Set precision to 6 decimal places std::cout << "Points within " << radius << "mm of (" << target.x << ", " << target.y << ", " << target.z << "):" << std::endl; for (const auto& p : result) { std::cout << "(" << p.x << ", " << p.y << ", " << p.z << ")" << std::endl; } // Cleanup and other code... }
通过使用 `std::fixed` 和 `std::setprecision(6)`,我设置了输出的精度为 6 位小数,确保了打印结果的小数位数与原始数据一致。
python的实现(作为验证AI的答案是否正确)
from scipy.spatial import KDTree import time def read_points_from_file(filename): points = [] with open(filename, 'r') as file: for line in file: # 解析每行数据,并转换为浮点数 x, y, z = map(float, line.strip().split(',')) points.append((x, y, z)) return points def find_points_in_range(kd_tree, target_point, radius): # 使用 KD 树的 query_ball_point 方法搜索指定范围内的所有点 indices = kd_tree.query_ball_point(target_point, radius) return [kd_tree.data[i] for i in indices] def main(): start_time = time.time() #filename = "D:/test_pro/old/CppTest/算法测试/test0/x64/Debug/test.txt"; filename = "D:/test_pro/old/CppTest/算法测试/test0/x64/Debug/20240502.txt"; points = read_points_from_file(filename) end_time = time.time() duration = (end_time - start_time) * 1000 # 将秒转换为毫秒 print("文件读取时间:", duration, "milliseconds") start_time = time.time() kd_tree = KDTree(points) end_time = time.time() duration = (end_time - start_time) * 1000 # 将秒转换为毫秒 print("构建kd树时间:", duration, "milliseconds") target_point = ( -141.874398,25.022764,2.484327) radius = 1 start_time = time.time() points_in_range = find_points_in_range(kd_tree, target_point, radius) end_time = time.time() duration = (end_time - start_time) * 1000 # 将秒转换为毫秒 print("查找时间:", duration, "milliseconds") print("一共有:", len(points_in_range), "个点") print("在指定范围内的点有:", points_in_range) for item in points_in_range: print(f'{item[0]},{item[1]},{item[2]}') if __name__ == "__main__": main()
小结
让我比较震惊的一点是,它无法直接给出正确的答案,但是如果能正确的指出它的错误,它居然就能一次次逼近正确,最终给出正确的答案。
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