无法使用opencv使用霍夫变换定位线

Posted 2023-02-21

【中文标题】无法使用opencv使用霍夫变换定位线

【英文标题】：Unable to locate the lines using hough transform using opencv

【发布时间】：2012-09-03 16:13:15

【问题描述】：

我想使用霍夫变换在下图中找到线条，但我失败了。谁能告诉我问题出在哪里？

我正在使用来自 opencv 的标准代码。

我正在使用 python 和 opencv 2.4.2

调查结果：

这些矩形线条非常参差不齐 边缘检测发现断边 即使您指定要连接的参数来填补空白也无济于事。

非常感谢。

编辑 正如“jpa”所建议的那样，图像被反转并且边缘检测步骤也被跳过

这是反转后的图像

使用的参数如下

    HoughLinesP(image,10, math.pi/2 ,10       ,None ,1000,          1)
    HoughLinesP(image,rho, theta    ,threshold,lines,minLineLength, maxLineGap)

输出如下，红色表示存在线条。

【问题讨论】：

对于这样的图像，跳过边缘检测步骤可能会带来更好的运气。只需反转图像（在黑底白字）并将其输入霍夫变换。另一方面，霍夫变换应该可以很好地处理破损的边缘。您可以发布霍夫变换的输出图像吗？

您好，谢谢您的回答，我已经跳过了边缘检测并反转了图像，我正在发布结果图像和参数，以便您知道，atm，行太多，大约 1000以及有白色的部分。

好的；使用自适应阈值，您可能可以摆脱白色背景。

也许您应该将 Python 标记添加到您的问题中。

【参考方案1】：

将您的原始图像作为以下程序的输入会产生以下结果：

绿线表示成功检测到的内容。该程序是对 OpenCV 附带的原始 squares 示例稍作修改。

由你来编写忽略最大行（identify the paper）的代码。

这些行存储在vector<vector<Point> > squares 中声明的main()：

#include "highgui.h"
#include "cv.h"

#include <iostream>
#include <math.h>
#include <string.h>

using namespace cv;
using namespace std;

void help()

        cout <<
        "\nA program using pyramid scaling, Canny, contours, contour simpification and\n"
        "memory storage (it's got it all folks) to find\n"
        "squares in a list of images pic1-6.png\n"
        "Returns sequence of squares detected on the image.\n"
        "the sequence is stored in the specified memory storage\n"
        "Call:\n"
        "./squares\n"
    "Using OpenCV version %s\n" << CV_VERSION << "\n" << endl;



int thresh = 70, N = 2; 
const char* wndname = "Square Detection Demo";

// helper function:
// finds a cosine of angle between vectors
// from pt0->pt1 and from pt0->pt2
double angle( Point pt1, Point pt2, Point pt0 )

    double dx1 = pt1.x - pt0.x;
    double dy1 = pt1.y - pt0.y;
    double dx2 = pt2.x - pt0.x;
    double dy2 = pt2.y - pt0.y;
    return (dx1*dx2 + dy1*dy2)/sqrt((dx1*dx1 + dy1*dy1)*(dx2*dx2 + dy2*dy2) + 1e-10);


// returns sequence of squares detected on the image.
// the sequence is stored in the specified memory storage
void findSquares( const Mat& image, vector<vector<Point> >& squares )

    squares.clear();

    Mat pyr, timg, gray0(image.size(), CV_8U), gray;

    // karlphillip: dilate the image so this technique can detect the white square,
    Mat out(image);
    dilate(out, out, Mat(), Point(-1,-1));
    // then blur it so that the ocean/sea become one big segment to avoid detecting them as 2 big squares.
    medianBlur(out, out, 3);

    // down-scale and upscale the image to filter out the noise
    pyrDown(out, pyr, Size(out.cols/2, out.rows/2));
    pyrUp(pyr, timg, out.size());
    vector<vector<Point> > contours;

    // find squares in every color plane of the image
    for( int c = 0; c < 1; 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:
                //     tgray(x,y) = gray(x,y) < (l+1)*255/N ? 255 : 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++ )
            
                // approximate contour with accuracy proportional
                // to the contour perimeter
                approxPolyDP(Mat(contours[i]), approx, arcLength(Mat(contours[i]), true)*0.02, true);

                // square contours should have 4 vertices after approximation
                // relatively large area (to filter out noisy contours)
                // and be convex.
                // Note: absolute value of an area is used because
                // area may be positive or negative - in accordance with the
                // contour orientation
                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 cosines of all angles are small
                    // (all angles are ~90 degree) then write quandrange
                    // vertices to resultant sequence
                    if( maxCosine < 0.3 )
                        squares.push_back(approx);
                
            
        
    
    

// the function draws all the squares in the image
void drawSquares( Mat& image, const vector<vector<Point> >& squares )

    for( size_t i = 1; i < squares.size(); i++ )
    
        const Point* p = &squares[i][0];
        int n = (int)squares[i].size();
        polylines(image, &p, &n, 1, true, Scalar(0,255,0), 3, CV_AA);
    

    imshow(wndname, image);
    

int main(int argc, char** argv)

    if (argc < 2)
    
        cout << "Usage: ./program <file>" << endl;
        return -1;
    

    static const char* names[] =  argv[1], 0 ;

    help();
    namedWindow( wndname, 1 );
    vector<vector<Point> > squares;

    for( int i = 0; names[i] != 0; i++ )
    
        Mat image = imread(names[i], 1);
        if( image.empty() )
        
            cout << "Couldn't load " << names[i] << endl;
            continue;
        

        findSquares(image, squares);
        drawSquares(image, squares);
        imwrite("out.jpg", image);

        int c = waitKey();
        if( (char)c == 27 )
            break;
    

    return 0;

【讨论】：

@karlphilip，非常感谢您的回答，看起来很酷。但问题是它找到了边角为 90 度的正方形。我正在处理的具体问题是检测原始形式的边并将它们优化为 90 度。用正方形近似轮廓会失去这个。有什么建议吗？

你能告诉我在这里使用 mixChannels 的确切目的是什么，我假设只是复制数据 mixChannels(&timg, 1, &gray0, 1, ch, 1);谢谢

更多信息：mixChannels()

关于您的第一个问题，您是否使用角上具有不同角度（不是 90 度）的正方形测试了此应用程序？