从边缘检测器输出修剪短线段？

Question

我正在寻找一种算法来修剪边缘检测器输出中的短线段。从下面的图像（和链接）中可以看出，检测到的几个小边缘不是“长”线。理想情况下，我希望加工后只显示四边形的四边，但如果有几条偏离线，那就没什么大不了的......有什么建议吗？

Image Link

Answer 1

在找到边缘之前，使用打开或关闭操作（或两者）预处理图像，即侵蚀，然后 dilate 或 dilate ，然后是 erode 。这应该删除较小的对象，但保留较大的对象大致相同。

我已经找到了在线示例，我能找到的最好的是this PDF的第41页。

Answer 2

我怀疑这可以通过简单的本地操作来完成。看看你想要保留的矩形 - 有几个间隙，因此执行局部操作来移除短线段可能会大大降低所需输出的质量。

因此，我会尝试通过关闭间隙，拟合多边形或类似的东西来检测矩形作为重要内容，然后在第二步中丢弃剩余的不重要内容。可能是Hough transform可以提供帮助。

<强>更新

我刚刚使用了这个sample application使用内核Hough变换和你的样本图像，并得到了四条适合你矩形的漂亮线条。

Answer 3

如果有人踩到这个帖子，OpenCV 2.x会带来一个名为 squares.cpp 的例子，基本上可以解决这个问题。

我对应用程序稍作修改以改进 quadrangle 的检测

<强>代码：

#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 Demonized";

// 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 only in the first color plane
    for( int c = 0; c < 1; c++ ) // was: c < 3
    {
        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;
}

Answer 4

霍夫变换可能是一项非常昂贵的操作。

可以在您的案例中正常运作的替代方案如下：

1. 运行2个数学形态学运算，分别称为图像闭合（http://homepages.inf.ed.ac.uk/rbf/HIPR2/close.htm），其水平和垂直线（由测试确定的给定长度）结构元素。关键是要关闭大矩形中的所有间隙。

2. 运行连通组件分析。如果您已经有效地完成了形态学，那么大矩形将作为一个连接组件出现。然后它只是迭代遍历所有连接的组件，并挑选出应该是大矩形的最可能的候选者。

Answer 5

可能找到连接的组件，然后移除小于X像素的组件（根据经验确定），然后沿水平/垂直线扩张以重新连接矩形内的间隙

Answer 6

可以遵循两种主要技术：

1. 基于矢量的操作：将像素岛映射到簇（blob，voronoi区域等）。然后应用一些启发式方法来纠正细分，如Teh-Chin链式近似算法，并对矢量元素（起点，终点，长度，方向等）进行修剪。

2. 基于设置的操作：对数据进行聚类（如上所述）。对于每个聚类，通过查找仅显示1个有意义特征值的聚类（或者如果查找“胖”段，则可以类似于省略号），计算主要成分并检测圆形或任何其他形状的线条。检查与特征值相关的特征向量，以获得有关斑点方向的信息，并做出选择。

使用OpenCV可以很容易地探索这两种方式（前者确实属于算法的“轮廓分析”类别）。

Answer 7

这是一个简单的形态过滤解决方案，遵循@ Tom10：

matlab中的解决方案：

se1 = strel('line',5,180);            % linear horizontal structuring element 
se2 = strel('line',5,90);             % linear vertical structuring element 
I = rgb2gray(imread('test.jpg'))>80;  % threshold (since i had a grayscale version of the image)
Idil = imdilate(imdilate(I,se1),se2); % dilate contours so that they connect
Idil_area = bwareaopen(Idil,1200);    % area filter them to remove the small components

这个想法是基本上连接水平轮廓以制作一个大的组件，然后通过区域打开过滤器来过滤以获得矩形。

<强>结果：