找到论文的一角 [英] Find corner of papers

问题描述

我是openCV的新手，因此在过去的3到4天里一直在努力，我已经检测到纸张边界，现在我想在角落画出4个圆圈。

I am new in openCV so struggling from last 3 to 4 days, I already detect paper sheet boundary, Now I wants to draw 4 circles on the corners.

我从这段代码中画出边界

I draw boundary from this code

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，所以在我看来我有左上角点p-> x和p-> y，但是我如何得到其他角落，我在这个折线方法的参数& n中也很困惑，这个折线方法如何绘制完整的矩形？

I am new in openCV, So In my opinion I have upper left corner points p->x and p->y, But how I get the others corners, I am also confused in parameter &n in this polylines method, how this polylines method draw complete rectangle?

当我使用边界矩形时，它并不完美，它在纸张的一侧留下了一点空间。

When I use bounding rect, It's not perfect it give's little space on side of paper sheet.

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

推荐答案

参考我的原始代码，它只是检测图像上的方块。

Take for reference my original code, which simply detects squares on an image.

这意味着在应用程序的 main 方法中，您将编写类似下面的伪代码来调用 find_squares（）：

That means that in the main method of the application you would write something like the following pseudo-code to call 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;

这个技巧依赖于 find_largest_square（）下面：

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];
}

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