OpenCV学习三十三：pointPolygonTest 检测点是否在轮廓内

Posted 2022-08-16 Thomas会写字

opencv函数

pointPolygonTest：

C++: double pointPolygonTest(InputArray contour, Point2f pt, bool measureDist)

用于测试一个点是否在多边形中
当measureDist设置为true时，返回实际距离值。若返回值为正，表示点在多边形内部，返回值为负，表示在多边形外部，返回值为0，表示在多边形上。
当measureDist设置为false时，返回 -1、0、1三个固定值。若返回值为+1，表示点在多边形内部，返回值为-1，表示在多边形外部，返回值为0，表示在多边形上。

#include <opencv2/opencv.hpp>
#include "iostream"

using namespace std;
using namespace cv;

int main( int argc, char** argv )

	Mat img = imread("fengye.jpg");
	threshold(img, img, 127, 255, CV_THRESH_BINARY);
	cvtColor(img ,img ,CV_BGR2GRAY);
	
	vector<vector<Point>> contours;
	vector<Vec4i> hierachy;
	findContours(img, contours, hierachy, RETR_TREE, CV_CHAIN_APPROX_SIMPLE, Point(0, 0));
	drawContours(img, contours, 1, Scalar::all(255), 2, 8, hierachy);

	//这里说明下，轮廓1是圈出的枫叶
	Mat dist = Mat::zeros(img.size(), CV_32FC1);
	for (int i=0; i<img.rows; i++)	
		for (int j=0; j<img.cols; j++)	
			double distance = pointPolygonTest(contours[1], Point2f(j, i), true);
			dist.at<float>(i, j) = static_cast<float>(distance);
		
	



	double minVal, maxVal;
	minMaxLoc(dist, &minVal, &maxVal, 0, 0, Mat());
	Mat drawImg1 = Mat::zeros(img.size(), CV_8UC3);
	int thredVal1 = 1;
	for (int i=0; i<img.rows; i++)	
		for (int j=0; j<img.cols; j++)	
			double distance = dist.at<float>(i,j);
			if (distance > thredVal1)
				drawImg1.at<Vec3b>(i,j)[0] = (uchar)(abs(distance/maxVal)*255);
			 else if (distance < (thredVal1 *(-1)) )
				drawImg1.at<Vec3b>(i,j)[2] = (uchar)(abs(distance/minVal)*255);
			 else 
				drawImg1.at<Vec3b>(i,j)[0] = (uchar)(abs(255-distance));
				drawImg1.at<Vec3b>(i,j)[1] = (uchar)(abs(255-distance));
				drawImg1.at<Vec3b>(i,j)[2] = (uchar)(abs(255-distance));
			
		
	
	imshow("drawImg1", drawImg1);
	imwrite("drawImg1.jpg", drawImg1);



	normalize(dist, dist, -255, 255, NORM_MINMAX);
	Mat drawImg2 = Mat::zeros(img.size(), CV_8UC3);
	int thredVal2 = 3;
	for (int i=0; i<img.rows; i++)	
		for (int j=0; j<img.cols; j++)	
			double distance = dist.at<float>(i,j);
			if (distance > thredVal2)
				drawImg2.at<Vec3b>(i,j)[0] = (uchar)(distance);
			 else if (distance < (thredVal2 *(-1)) )
				drawImg2.at<Vec3b>(i,j)[2] = (uchar)(distance*(-1));
			 else 
				drawImg2.at<Vec3b>(i,j)[0] = (uchar)(255);
				drawImg2.at<Vec3b>(i,j)[1] = (uchar)(255);
				drawImg2.at<Vec3b>(i,j)[2] = (uchar)(255);
			
		
	
	imshow("drawImg2", drawImg2);
	imwrite("drawImg2.jpg", drawImg2);
	
	imshow("img", img);
		
	waitKey();
	return 0;

原图

 

drawImg1

 

drawImg2

说明下：白色线的位置差异是由于归一化的方式不同，在第一种方式中肯定可以锁定到轮廓上，在第二种方式中由于最大最小距离的绝对值不同。