关于houghlines函数角度问题的说明

以上是opecv reference里面的说明。

Image必须是8位单通道图（可以使灰度图、二值图、边缘图等）

Rho：距离分辨率，一般为1

Theta：角度分辨率，一般为CV_PI/180

Threshold：阈值，只返回像素和大于threshold的直线

Srn：（猜测）距离缩放

Stn：（猜测）角度缩放

重点说明

Lines：

—p: 直线到原点（左上角）的距离

—：角度

这里的角度文档上有写是vertical line~0°，horizontal line~90°但是它没说方向是顺时针啊！

这导致我不知道下面这个角度到底是怎么计算的

An example using the Hough line detector can be found at opencv_source_code/samples/cpp/houghlines.cpp

大家可以再画图里画两条直线验证

#include "opencv2/highgui/highgui.hpp"
#include "opencv2/imgproc/imgproc.hpp"

#include <iostream>

using namespace cv;
using namespace std;

static void help()
{
    cout << "\nThis program demonstrates line finding with the Hough transform.\n"
        "Usage:\n"
        "./houghlines <image_name>, Default is pic1.png\n" << endl;
}

int main(int argc, char** argv)
{
    const char* filename = argc >= 2 ? argv[1] : "d:/vs/cardPic/w.png";

    Mat src = imread(filename, 0);
    if(src.empty())
    {
        help();
        cout << "can not open " << filename << endl;
        return -1;
    }

    Mat dst, cdst;
    Canny(src, dst, 50, 200, 3);
    cvtColor(dst, cdst, COLOR_GRAY2BGR);

    vector<Vec2f> lines;
    HoughLines(dst, lines, 1, CV_PI/180, 80, 0, 0 );

    for( size_t i = 0; i < lines.size(); i++ )
    {
        float rho = lines[i][0], theta = lines[i][1];
        cout<<i<<":"<<theta*180/CV_PI<<endl;
        Point pt1, pt2;
        double a = cos(theta), b = sin(theta);
        double x0 = a*rho, y0 = b*rho;
        pt1.x = cvRound(x0 + 1000*(-b));
        pt1.y = cvRound(y0 + 1000*(a));
        pt2.x = cvRound(x0 - 1000*(-b));
        pt2.y = cvRound(y0 - 1000*(a));
        line( cdst, pt1, pt2, Scalar(0,0,255), 1, CV_AA);
    }

    imshow("source", src);
    imshow("detected lines", cdst);

    waitKey();

    return 0;
}