我需要从网络摄像头绘制带有2个最大对象的矩形。我已经用网络摄像头中的2个最大的对象绘制轮廓但现在我混淆了如何绘制2个最大的矩形。 有人可以给我看一下代码Please~
//find and draw contours
void showconvex(Mat &thresh,Mat &frame) {
int largestIndex = 0;
int largestContour = 0;
int secondLargestIndex = 0;
int secondLargestContour = 0;
vector<vector<Point> > contours;
vector<Vec4i> hierarchy;
//find contours
findContours(thresh, contours, hierarchy, CV_RETR_TREE, CV_CHAIN_APPROX_SIMPLE);
/// Find the convex hull object for each contour
vector<vector<Point> >hull(contours.size());
vector<vector<int> >inthull(contours.size());
vector<vector<Vec4i> >defects(contours.size());
for (int i = 0; i < contours.size(); i++)
{
convexHull(Mat(contours[i]), hull[i], false);
convexHull(Mat(contours[i]),inthull[i], false);
if (inthull[i].size()>3)
convexityDefects(contours[i], inthull[i], defects[i]);
}
//find 2 largest contour
for( int i = 0; i< contours.size(); i++ )
{
if(contours[i].size() > largestContour)
{
secondLargestContour = largestContour;
secondLargestIndex = largestIndex;
largestContour = contours[i].size();
largestIndex = i;
}
else if(contours[i].size() > secondLargestContour)
{
secondLargestContour = contours[i].size();
secondLargestIndex = i;
}
}
//show contours of 2 biggest and hull as well
if(contours.size()>0)
{
//check for contouraea function if error occur
//draw the 2 largest contour using previously stored index.
drawContours(frame, contours, largestIndex, CV_RGB(0,255,0), 2, 8, hierarchy);
drawContours(frame, contours, secondLargestIndex, CV_RGB(0,255,0), 2, 8, hierarchy);
}
}
答案 0 :(得分:1)
看看下面的代码
基于按边界框或区域排序轮廓。
#include "opencv2/imgproc.hpp"
#include "opencv2/highgui.hpp"
using namespace cv;
using namespace std;
struct contour_sorter_dsc // sorts contours by their bounding boxes descending
{
bool operator ()( const vector<Point>& a, const vector<Point> & b )
{
Rect ra( boundingRect(a) );
Rect rb( boundingRect(b) );
return ( ( rb.width * rb.height ) < ( ra.width * ra.height ) );
}
};
struct contour_sorter_dsc_area // sorts contours by their areas descending
{
bool operator ()( const vector<Point>& a, const vector<Point> & b )
{
double area_a = contourArea( a );
double area_b = contourArea( b );
return ( area_b < area_a );
}
};
int main( int argc, char** argv )
{
Mat src = imread( argv[1] );
if( src.empty() )
{
return -1;
}
Mat canvas1 = src.clone();
Mat canvas2 = src.clone();
Mat gray;
cvtColor( src, gray, COLOR_BGR2GRAY );
gray = gray > 127; // binarize image
vector<vector<Point> > contours;
findContours( gray, contours, RETR_LIST, CHAIN_APPROX_SIMPLE );
sort(contours.begin(), contours.end(), contour_sorter_dsc());
for( size_t i = 0; i< 2; i++ )
{ // checks if the first contour is image boundary
if( contours[0][0] == Point( 1, 1 ) & contours[0][1] == Point( 1, gray.rows -2 )
& contours[0][2] == Point( gray.cols - 2, gray.rows -2 ) & contours[0][3] == Point( gray.cols - 2, 1 ) )
{
contours[0] = contours[1];
contours[1] = contours[2];
}
if( i < contours.size())
{
drawContours( canvas1, contours, i, Scalar( 255,255,0 ) );
Rect minRect = boundingRect( Mat(contours[i]) );
rectangle( canvas1, minRect, Scalar( 0, 0, 255 ) );
}
}
imshow( "result of sorting contours by bounding boxes ", canvas1 );
sort(contours.begin(), contours.end(), contour_sorter_dsc_area());
for( size_t i = 0; i< 2; i++ )
{ // checks if the first contour is image boundary
if( contours[0][0] == Point( 1, 1 ) & contours[0][1] == Point( 1, gray.rows -2 )
& contours[0][2] == Point( gray.cols - 2, gray.rows -2 ) & contours[0][3] == Point( gray.cols - 2, 1 ) )
{
contours[0] = contours[1];
contours[1] = contours[2];
}
if( i < contours.size())
{
drawContours( canvas2, contours, i, Scalar( 255,255,0 ) );
Rect minRect = boundingRect( Mat(contours[i]) );
rectangle( canvas2, minRect, Scalar( 0, 0, 255 ) );
}
}
imshow( "result of sorting contours by areas ", canvas2 );
waitKey();
return 0;
}
输入图片
根据排序类型
的结果图像