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我在 OpenCV 中使用霍夫变换来检测线条。但是，我事先知道我只需要在非常有限的角度范围内（大约 10 度左右）内的线条。我在对性能非常敏感的设置中执行此操作，因此我想避免在检测其他角度的线条上花费额外的工作，这些线条我事先知道我不关心。

我可以从 OpenCV 中提取 Hough 源代码，然后将其破解以获取 min_rho 和 max_rho 参数，但我想要一种不那么脆弱的方法（必须手动更新我的代码，每次 OpenCV 更新等）。

这里最好的方法是什么？

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3 回答 3

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If you use the Probabilistic Hough transform then the output is in the form of a cvPoint each for lines[0] and lines[1] parameters. We can get x and y co-ordinated for each of the two points by pt1.x, pt1.y and pt2.x and pt2.y. Then use the simple formula for finding slope of a line - (y2-y1)/(x2-x1). Taking arctan (tan inverse) of that will yield that angle in radians. Then simply filter out desired angles from the values for each hough line obtained.

于 2011-02-05T16:19:04.777 回答
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好吧，我已经修改了icvHoughlines函数以适应一定的角度范围。我确信也有更简洁的方式来处理内存分配，但是对于从 180 度到 60 度的角度范围，我的速度提升从 100 毫秒到 33 毫秒，所以我对此很满意。

请注意，此代码还输出累加器值。另外，我只输出 1 行，因为这符合我的目的，但那里并没​​有真正的收获。

static void
icvHoughLinesStandard2( const CvMat* img, float rho, float theta,
                       int threshold, CvSeq *lines, int linesMax )
{
    cv::AutoBuffer<int> _accum, _sort_buf;
    cv::AutoBuffer<float> _tabSin, _tabCos;

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

    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) );

    // find n and ang limits (in our case we want 60 to 120
    float limit_min = 60.0/180.0*PI;
    float limit_max = 120.0/180.0*PI;

    //num_steps = (limit_max - limit_min)/theta;
    int start_n = floor(limit_min/theta);
    int stop_n = floor(limit_max/theta);

    for( ang = limit_min, n = start_n; n < stop_n; 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 )
                        //
        for( n = start_n; n < stop_n; n++ )
                {
                    r = cvRound( j * tabCos[n] + i * tabSin[n] );
                    r += (numrho - 1) / 2;
                    accum[(n+1) * (numrho+2) + r+1]++;
                }
        }



    int max_accum = 0;
    int max_ind = 0;

    for( r = 0; r < numrho; r++ )
    {
        for( n = start_n; n < stop_n; n++ )
        {
            int base = (n+1) * (numrho+2) + r+1;
            if (accum[base] > max_accum)
            {
                max_accum = accum[base];
                max_ind = base;
            }
        }
    }   

    CvLinePolar2 line;
    scale = 1./(numrho+2);
    int idx = max_ind;
    n = cvFloor(idx*scale) - 1;
    r = idx - (n+1)*(numrho+2) - 1;
    line.rho = (r - (numrho - 1)*0.5f) * rho;
    line.angle = n * theta;
    line.votes = accum[idx];
    cvSeqPush( lines, &line );

}
于 2010-11-03T16:11:05.820 回答
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我认为使用标准 HoughLines(...) 函数更自然，它直接在 rho 和 theta 项中提供线的集合并从中选择必要的角度范围，而不是从段端点重新计算角度。

于 2015-01-19T15:56:58.607 回答