11

我是 openCV 的新手，过去 3 到 4 天都在苦苦挣扎，我已经检测到纸张边界，现在我想在角落画 4 个圆圈。

我从这段代码中画出边界

const cv::Point* p = &squares[i][0];

int n = (int)squares[i].size();

polylines(image, &p,&n, 1, true, Scalar(255,255,0), 5, CV_AA);

我是openCV的新手，所以我认为我有左上角点p-> x和p-> y，但是我如何得到其他角，我也对这个折线方法中的参数&n感到困惑，这个折线方法如何绘制完整的矩形？

当我使用边界矩形时，它并不完美，它在纸张侧面留下了很小的空间。

非常感谢任何帮助

代码是：

- (cv::Mat)finshWork:(cv::Mat &)image
{
// read in the apple (change path to the file)
Mat img0 =image;// imread("/home/philipp/img/apple.jpg", 1);

Mat img1;
cvtColor(img0, img1, CV_RGB2GRAY);

// apply your filter
Canny(img1, img1, 100, 200);

// find the contours
vector< vector<cv::Point> > contours;
findContours(img1, contours, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_NONE);

/////for SQUARE CODE
std::vector<std::vector<cv::Point> > squares;
std::vector<cv::Point> approx;
for( size_t i = 0; i < contours.size(); i++ )
{
    cv::approxPolyDP(cv::Mat(contours[i]), approx, arcLength(cv::Mat(contours[i]), true)*0.02, true);
    if( approx.size() == 4 && fabs(contourArea(cv::Mat(approx))) > 1000 && cv::isContourConvex(cv::Mat(approx))) {
        double maxCosine = 0;

        for( int j = 2; j < 5; j++ )
        {
            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);
            cv::Point newPoint = approx[0]; 

            NSLog(@"x is %d and  y is %d",newPoint.x,newPoint.y);
        }
    }
}

const cv::Point* p = &squares[0][0];


int n = (int)squares[0].size();

NSLog(@"%d",n);


//THIS IS WORKING CODE              

    polylines(image, &p,&n, 1, true, Scalar(0,0,255), 10, CV_AA);
    //polylines(image, &p,&n, 1, true, Scalar(255,255,0), 5, CV_AA);
////////////
}

谢谢

4

1 回答 1

20

参考我的原始代码，它只是检测图像上的正方形。

这意味着在应用程序的main方法中，您将编写类似于以下伪代码的内容来调用find_squares()：

Mat image = imread("test.jpg", 1);

// Detect all regions in the image that are similar to a rectangle
vector<vector<Point> > squares;
find_squares(image, squares);

// The largest of them probably represents the paper
vector<Point> largest_square;
find_largest_square(squares, largest_square);

// Print the x,y coordinates of the square
cout << "Point 1: " << largest_square[0] << endl;
cout << "Point 2: " << largest_square[1] << endl;
cout << "Point 3: " << largest_square[2] << endl;
cout << "Point 4: " << largest_square[3] << endl;

诀窍依赖于find_largest_square()下面介绍：

void find_largest_square(const vector<vector<Point> >& squares, vector<Point>& biggest_square)
{
    if (!squares.size())
    {
            // no squares detected
            return;
    }

    int max_width = 0;
    int max_height = 0;
    int max_square_idx = 0;
    const int n_points = 4;

    for (size_t i = 0; i < squares.size(); i++)
    {
            // Convert a set of 4 unordered Points into a meaningful cv::Rect structure.
            Rect rectangle = boundingRect(Mat(squares[i]));

    //        cout << "find_largest_square: #" << i << " rectangle x:" << rectangle.x << " y:" << rectangle.y << " " << rectangle.width << "x" << rectangle.height << endl;

            // Store the index position of the biggest square found
            if ((rectangle.width >= max_width) && (rectangle.height >= max_height))
            {
                    max_width = rectangle.width;
                    max_height = rectangle.height;
                    max_square_idx = i;
            }
    }

    biggest_square = squares[max_square_idx];
}
于 2012-11-23T16:28:57.567 回答