我在图像中有点。我需要检测最共线的点。最快的方法是使用霍夫变换，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，还需要一个容差阈值来检测几乎检测到的点。有人可以帮助调整这个阈值吗？我需要至少四个半共线点来检测它们所属的线。
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
- 第三个点被三行检测到
摆脱重叠留置权的最佳方法是什么？或者如何调整容差阈值以仅通过一条线检测半共线点？
这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);
}
}