这是方块检测示例的输出我的问题是过滤这个方块
另一个问题是我必须只取除所有图像之外的最大对象。
下面是检测代码:
static void findSquares( const Mat& image, vector >& squares ){
squares.clear();
Mat pyr, timg, gray0(image.size(), CV_8U), gray;
// down-scale and upscale the image to filter out the noise
pyrDown(image, pyr, Size(image.cols/2, image.rows/2));
pyrUp(pyr, timg, image.size());
vector<vector<Point> > contours;
// find squares in every color plane of the image
for( int c = 0; c < 3; c++ )
{
int ch[] = {c, 0};
mixChannels(&timg, 1, &gray0, 1, ch, 1);
// try several threshold levels
for( int l = 0; l < N; l++ )
{
// hack: use Canny instead of zero threshold level.
// Canny helps to catch squares with gradient shading
if( l == 0 )
{
// apply Canny. Take the upper threshold from slider
// and set the lower to 0 (which forces edges merging)
Canny(gray0, gray, 0, thresh, 5);
// dilate canny output to remove potential
// holes between edge segments
dilate(gray, gray, Mat(), Point(-1,-1));
}
else
{
// apply threshold if l!=0:
gray = gray0 >= (l+1)*255/N;
}
// find contours and store them all as a list
findContours(gray, contours, CV_RETR_LIST, CV_CHAIN_APPROX_SIMPLE);
vector<Point> approx;
// test each contour
for( size_t i = 0; i < contours.size(); i++ )
{
approxPolyDP(Mat(contours[i]), approx, arcLength(Mat(contours[i]), true)*0.02, true);
if( approx.size() == 4 &&
fabs(contourArea(Mat(approx))) > 1000 &&
isContourConvex(Mat(approx)) )
{
double maxCosine = 0;
for( int j = 2; j < 5; j++ )
{
// find the maximum cosine of the angle between joint edges
double cosine = fabs(angle(approx[j%4], approx[j-2], approx[j-1]));
maxCosine = MAX(maxCosine, cosine);
}
if( maxCosine < 0.3 )
squares.push_back(approx);
}
}
}
}
}
最佳答案
您需要查看 findContours() 的标志。您可以设置一个名为 CV_RETR_EXTERNAL 的标志,它将仅返回最外层的轮廓(其中的所有轮廓都将被丢弃)。这可能会返回整个帧,因此您需要缩小搜索范围,以便它不会检查您的帧边界。使用函数 copyMakeBorder() 来完成此操作。我还建议删除您的 dilate 函数,因为它可能会导致线条两侧出现重复的轮廓(如果您删除了 dilate,您甚至可能不需要边框)。这是我的输出:
关于OpenCV 平方 : filtering output,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/14975304/