寻找包裹轮廓的最小正矩形:boundingRect 函数
返回矩阵应满足:① 轮廓上的点均在矩阵空间内。② 矩阵是正矩阵(矩形的边界与图像边界平行)。
Rect boundingRect(InputArray points);
- 唯一一个参数是输入的二维点集,可以是 vector 或 Mat 类型。
代码示例:
#include<opencv.hpp> #include<iostream>
using namespace cv; using namespace std; int main(){ Mat src = imread("C:/Users/齐明洋/Desktop/7.jpg"); imshow("src", src); Mat gray, bin_img; cvtColor(src, gray, COLOR_BGR2GRAY); //将原图转换为灰度图
imshow("gray", gray); //二值化
threshold(gray, bin_img, 150, 255, THRESH_BINARY_INV); imshow("bin_img", bin_img); //寻找最外围轮廓
vector<vector<Point> >contours; findContours(bin_img, contours, RETR_EXTERNAL, CHAIN_APPROX_NONE); //绘制边界矩阵
RNG rngs = { 12345 }; Mat dst = Mat::zeros(src.size(), src.type()); for (int i = 0; i < contours.size(); i++) { Scalar colors = Scalar(rngs.uniform(0, 255), rngs.uniform(0, 255), rngs.uniform(0, 255)); drawContours(dst, contours, i, colors, 1); Rect rects = boundingRect(contours[i]); rectangle(dst, rects, colors, 2); } imshow("dst", dst); waitKey(0); }
效果演示:
寻找包裹轮廓的最小斜矩形:minAreaRect 函数
返回矩阵应满足:① 轮廓上的点均在矩阵空间内。② 没有面积更小的满足条件的矩阵(与 boundingRect 返回结果的区别是:矩形的边界不必与图像边界平行)。
需要补充的是,求点集的拟合椭圆(fitEllipse() https://www.cnblogs.com/bjxqmy/p/12354750.html)便是求斜矩阵内最大的椭圆(矩阵长宽分别做椭圆长轴、短轴)。
RotatedRect minAreaRect(InputArray points);
- 唯一一个参数是输入的二维点集,可以是 vector 或 Mat 类型。
代码示例:
#include<opencv.hpp> #include<iostream> #include<vector>
using namespace cv; using namespace std; int main() { Mat src = imread("C:/Users/齐明洋/Desktop/7.jpg"); imshow("src", src); Mat gray, bin_img; cvtColor(src, gray, COLOR_BGR2GRAY); //将原图转换为灰度图
imshow("gray", gray); //二值化
threshold(gray, bin_img, 150, 255, THRESH_BINARY_INV); imshow("bin_img", bin_img); //寻找最外围轮廓
vector<vector<Point> >contours; findContours(bin_img, contours, RETR_EXTERNAL, CHAIN_APPROX_NONE); //绘制最小边界矩阵
RNG rngs = { 12345 }; Mat dst = Mat::zeros(src.size(), src.type()); Point2f pts[4]; for (int i = 0; i < contours.size(); i++) { Scalar colors = Scalar(rngs.uniform(0, 255), rngs.uniform(0, 255), rngs.uniform(0, 255)); drawContours(dst, contours, i, colors, 1);
RotatedRect rects = minAreaRect(contours[i]); rects.points(pts);//确定旋转矩阵的四个顶点 for (int i = 0; i < 4; i++) { line(dst, pts[i], pts[(i + 1) % 4], colors, 2); } } imshow("dst", dst); waitKey(0); }
效果演示:
借鉴博客:https://www.cnblogs.com/little-monkey/p/7429579.html
http://www.pianshen.com/article/4286104294/