opencv笔记（二十三）——OpenCV轮廓vector vector<Point>，vector<Vec4i>，vector<Rect>，vector<RotatedRect>

快来打我* 2022-05-15 09:38 882阅读 0赞

轮廓周围绘制矩形框

vector<vector<Point>>，vector<V e c 4 i>，vector<R e c t>，vector<R o t a t e d R e c t>

刚开始学OpenCV没多久遇到这些个东西不知道是什么，搞得很不舒服。

通过给轮廓绘制矩形框弄明白了这些东西。

**vector<vector<Point>>：**

vector容器里面放了一个vector容器，子容器里放点

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**vector<V e c 4 i>：**放了4维int向量

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**vector<R e ct>：** 像素width \* height   from  位置(x\*y)

![70 2][]

**vector<R o t a t e d R e c t>：**如图三个成员

![70 3][]

代码如下：

#include <opencv2/opencv.hpp> 
    #include <iostream>  
     
    using namespace cv;
    using namespace std;
     
    int main() {
    	Mat src, gray_src, drawImg, bin_output;
     
    	src = imread("F:/Temp_vs/pic/09.jpg");
    	namedWindow("input", CV_WINDOW_AUTOSIZE);
    	namedWindow("output", CV_WINDOW_AUTOSIZE);
    	imshow("input", src);
     
    	cvtColor(src, gray_src, CV_BGR2GRAY);
    	blur(gray_src, gray_src, Size(10, 10), Point(-1, -1), BORDER_DEFAULT);
     
    	//这些个类型
    	vector<vector<Point>> contours;
    	vector<Vec4i> hierarchy;
    	
    	threshold(gray_src, bin_output, 144, 255, 0);					//二值化
    	findContours(bin_output, contours, hierarchy, RETR_TREE, CHAIN_APPROX_SIMPLE, Point(0, 0));		//找轮廓
     
    	//这些个类型
    	vector<vector<Point>> contours_poly(contours.size());
    	vector<Rect> poly_rects(contours.size());
    	vector<RotatedRect> minRect(contours.size());
     
    	//取点
    	for (size_t i = 0; i < contours.size(); i++)
    	{
    		approxPolyDP(Mat(contours[i]), contours_poly[i], 3, true); //减少轮廓点数
     
    		poly_rects[i]=boundingRect(contours_poly[i]);//获取绘制矩形数据
     
    		if (contours_poly[i].size() > 5) {				
    			minRect[i] = minAreaRect(contours_poly[i]);//获取绘制旋转矩形数据
    		}
    	}
     
    	//开始绘制
    	src.copyTo(drawImg);
    	Point2f pst[4];//储存单个旋转矩形的四个点
    	cout << "----------Point2f pst[4]------------" << endl;
    	for (size_t i = 0; i < contours.size(); i++)
    	{
    		rectangle(drawImg, poly_rects[i],  Scalar(255, 0, 0), 2, 8);//绘制矩形框
     
    		minRect[i].points(pst);//用线段画矩形，将RotatedRect类型转化为四个点	
     
    		for (size_t u = 0; u < 4; u++)
    		{
    			line(drawImg, pst[u], pst[(u + 1) % 4], Scalar(0, 255, 0), 2, 8);
    			cout << pst[u];			//显示pst的数据
    		}
    		cout << endl;
    		Rect brect = minRect[i].boundingRect(); //返回包含旋转矩形的最小矩形  
    		rectangle(drawImg, brect,Scalar(0, 0, 255));
    	}
    	cout << endl;
    	imshow("output", drawImg);
     
    	cout << "----------vector<vector<Point>> contours_poly------------" << endl;
    	for (size_t i = 0; i < contours_poly.size(); i++)
    	{
    		cout << "第" << i << "行：";
    		for (size_t j = 0; j < contours_poly[i].size(); j++)
    		{
    			cout<<contours_poly[i][j];
    		}
    		cout << endl;
    	}
    	cout << endl;
     
    	cout << "----------vector<Vec4i> hierarchy------------" << endl;
    	for (size_t i = 0; i < hierarchy.size(); i++)
    	{
    		cout << hierarchy[i] << endl;
    	}
    	cout << endl;
    	
    	cout << "----------vector<Rect> poly_rects------------" << endl;
    	for (size_t i = 0; i < poly_rects.size(); i++)	
    	{
    		cout << poly_rects[i]<<endl;
    	}
    	cout << endl;
    	
    	cout << "---------vector<RotatedRect> minRect------------" << endl;
    	for (size_t i = 0; i <minRect.size(); i++)	//显示一下点minRect
    	{
    		cout <<"angle:"<< minRect[i].angle<<" center:"<< minRect[i].center<<" size:"<< minRect[i].size<<endl;
    	}
    	cout << endl;
     
    	waitKey(0);
    	return 0;
    }

![70 4][]

![70 5][]

转自：[https://blog.csdn.net/Ahuuua/article/details/80593388][https_blog.csdn.net_Ahuuua_article_details_80593388]

[70]: /images/20220515/3a1ee27c57404b4493df8aadb0aa5aae.png
[70 1]: /images/20220515/1c6b99e3477245f69d91acc9562a56bd.png
[70 2]: /images/20220515/ad4d7090eab64ddabd5e12fcd4b660cb.png
[70 3]: /images/20220515/50359cb81d3c4632a41f355689344d99.png
[70 4]: /images/20220515/4b84906a1895443aa2bc51855f21751a.png
[70 5]: /images/20220515/ef35c582e6aa494ebe588c417c82bc5d.png
[https_blog.csdn.net_Ahuuua_article_details_80593388]: https://blog.csdn.net/Ahuuua/article/details/80593388