如何使用OpenCV绘制特定边数的正多边形？

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OpenCV绘制正多边形的代码示例，提供详细参考，内容如下：

pythonimport cv2import numpy as np

def draw_polygon(center, radius, num_sides): angle=2 * np.pi / num_sides points=[] for i in range(num_sides): x=int(center[0] + radius * np.cos(i * angle)) y=int(center[1] + radius * np.sin(i * angle)) points.append((x, y)) return points

设置正多边形的中心点、半径和边数center=(200, 200)radius=100num_sides=5

生成正多边形的顶点坐标polygon_points=draw_polygon(center, radius, num_sides)

创建一个空白图像image=np.zeros((400, 400, 3), dtype=np.uint8)

绘制正多边形cv2.polylines(image, [np.array(polygon_points)], True, (255, 0, 0), 3)

显示图像cv2.imshow('Polygon', image)cv2.waitKey(0)cv2.destroyAllWindows()

OpenCV 绘制正多边形的具体代码，供大家参考，具体内容如下

#include <iostream> #include <opencv2\core\core.hpp> #include <opencv2\opencv.hpp> #include <opencv2\highgui\highgui.hpp> #include <opencv2\contrib\contrib.hpp> #include <fstream> using namespace cv; using namespace std; void DeleteRepetition(vector<Point> &Data) { vector<Point>::iterator it, it1; for (it = ++Data.begin(); it != Data.end();) { it1 = find(Data.begin(), it, *it); if (it1 != it) it = Data.erase(it); else it++; } } void Patterns(Mat *src, vector<Point> Dots, int fill) { DeleteRepetition(Dots); if (fill == -1) { Point *ImgDot = new Point(Dots.size()); for (int i = 0; i < Dots.size(); i++) { ImgDot[i] = Dots[i]; } const Point* ppt = ImgDot; int npt = Dots.size(); RNG &rng = theRNG(); Scalar color = Scalar(rng.uniform(100, 255), rng.uniform(100, 255), rng.uniform(100, 255)); cv::fillPoly(*src, &ppt, &npt, 1, color); } else { Dots.push_back(Dots[0]); RNG &rng = theRNG(); Scalar color = Scalar(rng.uniform(100, 255), rng.uniform(100, 255), rng.uniform(100, 255)); for (int i = 0; i < Dots.size() - 1; i++) { line(*src, Dots[i], Dots[i + 1], color, fill); } } } // www.w3cplus.com/canvas/drawing-regular-polygons.html // www.cnblogs.com/xcywt/p/9456526.html // 图像、中心点、半径、边数、旋转角度、线宽 void EquilateralPolygon(Mat *src, Point origin, int radius, int brim, int rotate, int fill) { if (brim < 3) return; if (rotate > 360) return; #define PI 3.14159265 #define ROTATE_COUNT 180 double nAgree = 360 / brim; // 计算旋转角度 double a = radius * cos(PI / brim); // 计算垂直向下的长度 double s = 2 * radius * sin(PI / brim); // 计算边长 vector<Point> Dots; Point D1, D2; D1.x = origin.x + radius*cos(-(((180 - nAgree) / 2) + rotate) * PI / 180); D1.y = origin.y - radius*sin(-(((180 - nAgree) / 2) + rotate) * PI / 180); D2.x = origin.x + radius*cos(-(((180 - nAgree) / 2) + nAgree + rotate) * PI / 180); D2.y = origin.y - radius*sin(-(((180 - nAgree) / 2) + nAgree + rotate) * PI / 180); // 第一条边的两个点 Dots.push_back(D1); Dots.push_back(D2); for (int i = 0; i < brim - 2; i++) { double dSinRot = sin((nAgree * (i + 1)) * PI / 180); double dCosRot = cos((nAgree * (i + 1)) * PI / 180); int x = origin.x + dCosRot * (D2.x - origin.x) - dSinRot * (D2.y - origin.y); int y = origin.y + dSinRot * (D2.x - origin.x) + dCosRot * (D2.y - origin.y); Dots.push_back(Point(x, y)); } Patterns(src, Dots, fill); Dots.clear(); } int main() { Mat Img = Mat::zeros(800, 800, CV_8UC3); Point O = Point(400, 400); circle(Img, O, 2, Scalar(0, 0, 255), -1); //中心点 EquilateralPolygon(&Img, O, 100, 3, 0, -1); // 填充的正三角形 EquilateralPolygon(&Img, O, 200, 3, 0, 1); // 不填充的正三角形 EquilateralPolygon(&Img, O, 200, 3, 30, 1); // 不填充的正三角形，顺时针旋转30度 EquilateralPolygon(&Img, O, 200, 3, 60, 1); // 不填充的正三角形，顺时针旋转60度 EquilateralPolygon(&Img, O, 200, 3, 90, 1); // 不填充的正三角形，顺时针旋转90度 EquilateralPolygon(&Img, O, 200, 3, 120, 1);// 不填充的正三角形，顺时针旋转120度 EquilateralPolygon(&Img, O, 200, 3, 150, 1);// 不填充的正三角形，顺时针旋转150度 EquilateralPolygon(&Img, O, 200, 3, 180, 1);// 不填充的正三角形，顺时针旋转180度 EquilateralPolygon(&Img, O, 230, 4, 0, 2); // 不填充的正四边形 EquilateralPolygon(&Img, O, 250, 5, 0, 3); // 不填充的正五边形 EquilateralPolygon(&Img, O, 270, 6, 0, 4); // 不填充的正六边形 EquilateralPolygon(&Img, O, 290, 7, 0, 5); // 不填充的正七边形 EquilateralPolygon(&Img, O, 310, 8, 0, 6); // 不填充的正八边形 EquilateralPolygon(&Img, O, 330, 9, 0, 7); // 不填充的正九边形 EquilateralPolygon(&Img, O, 350, 10, 0, 8);// 不填充的正十边形 imshow("正多边形", Img); waitKey(0); return 0; }

效果如下：

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