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这里是出放平方检测例的我的问题是过滤此正方形OpenCV的正方形:滤波输出
- 第一个问题是它的拉伸超过一人线相同的面积;
- 第二个是我只需要检测对象不是所有的图像。
另一个问题是我必须采取最大的对象,除了所有图像。
这里是检测代码:
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);
}
}
}
}
}
计算检测到的平方区域,然后取最大的一个。您可以尝试通过检查检测到的方块小于图像的95%来解除“整幅图像方块”。 – iiro 2013-02-20 09:05:30
也可以添加原始图像,以便用户可以对其进行操作并演示。 – 2013-02-20 09:05:34
原始图像 http://ozsulastik.com/p1.jpg http://ozsulastik.com/p2.jpg – 2013-02-20 09:11:49