Question

我在图像中有分数。我需要检测最共线的点。最快的方法是使用Hough变换，但我必须修改opencv方法。实际上我需要用检测到的线返回半共线点，因此我修改了极线结构。如图所示，还需要公差阈值来检测几乎检测到的点。有人可以帮助如何调整此阈值？我需要至少四个半共线点来检测它们所属的线。 ex 1 ex 2 ex 3

第一张图像的点由6条重叠线检测
中间图像的点没有被检测到
第三点通过三行检测

`![enter image description here

摆脱重叠留置权的最佳方法是什么？或者如何调整公差阈值以仅通过一条线检测半共线点？

是我自己的函数调用：

vector<CvLinePolar2> lines;
CvMat c_image = source1; // loaded image
HoughLinesStandard(&c_image,1,CV_PI/180,4,&lines,INT_MAX);

    typedef struct CvLinePolar2
    {
      float rho;
      float angle;
      vector<CvPoint> points;
    };

void HoughLinesStandard( const CvMat* img, float rho, float theta,
                   int threshold, vector<CvLinePolar2> *lines, int linesMax= INT_MAX )
{
cv::AutoBuffer<int> _accum, _sort_buf;
cv::AutoBuffer<float> _tabSin, _tabCos;


const uchar* image;
int step, width, height;
int numangle, numrho;
int total = 0;
int i, j;
float irho = 1 / rho;
double scale;

vector<vector<CvPoint>> lpoints;
CV_Assert( CV_IS_MAT(img) && CV_MAT_TYPE(img->type) == CV_8UC1 );

image = img->data.ptr;
step = img->step;
width = img->cols;
height = img->rows;

numangle = cvRound(CV_PI / theta);
numrho = cvRound(((width + height) * 2 + 1) / rho);

_accum.allocate((numangle+2) * (numrho+2));
_sort_buf.allocate(numangle * numrho);
_tabSin.allocate(numangle);
_tabCos.allocate(numangle);
int *accum = _accum, *sort_buf = _sort_buf;
float *tabSin = _tabSin, *tabCos = _tabCos;

memset( accum, 0, sizeof(accum[0]) * (numangle+2) * (numrho+2) );
//memset( lpoints, 0, sizeof(lpoints) );
lpoints.resize(sizeof(accum[0]) * (numangle+2) * (numrho+2));
float ang = 0;
for(int n = 0; n < numangle; ang += theta, n++ )
{
    tabSin[n] = (float)(sin(ang) * irho);
    tabCos[n] = (float)(cos(ang) * irho);
}

// stage 1. fill accumulator
for( i = 0; i < height; i++ )
    for( j = 0; j < width; j++ )
    {
        if( image[i * step + j] != 0 )
        {
            CvPoint pt;
            pt.x = j; pt.y = i;
            for(int n = 0; n < numangle; n++ )
            {
                int r = cvRound( j * tabCos[n] + i * tabSin[n] );
                r += (numrho - 1) / 2;

                int ind = (n+1) * (numrho+2) + r+1;
                int s = accum[ind];
                accum[ind]++;

                lpoints[ind].push_back(pt);


            }

        }
    }

// stage 2. find local maximums
for(int r = 0; r < numrho; r++ )
    for(int n = 0; n < numangle; n++ )
    {
        int base = (n+1) * (numrho+2) + r+1;
        if( accum[base] > threshold &&
            accum[base] > accum[base - 1] && accum[base] >= accum[base + 1] &&
            accum[base] > accum[base - numrho - 2] && accum[base] >= accum[base + numrho + 2] )
            sort_buf[total++] = base;
    }

// stage 3. sort the detected lines by accumulator value
icvHoughSortDescent32s( sort_buf, total, accum );

// stage 4. store the first min(total,linesMax) lines to the output buffer
linesMax = MIN(linesMax, total);
scale = 1./(numrho+2);
for( i = 0; i < linesMax; i++ )
{
    CvLinePolar2 line;
    int idx = sort_buf[i];
    int n = cvFloor(idx*scale) - 1;
    int r = idx - (n+1)*(numrho+2) - 1;
    line.rho = (r - (numrho - 1)*0.5f) * rho;
    line.angle = n * theta;

    line.points = lpoints[idx];

    lines->push_back(line);
}

}

Answer 1

一种方法是non-maximal suppression来稀疏潜在线的候选集。一旦确定了稀疏的潜在线，您就可以计算出满足某个角度或空间差异阈值的剩余线的平均值。

Answer 2

试试HoughLinesP .. opencv reference

霍夫变换返回共线和半共线点

2 个答案: