c++ - 使用opencv访问沿曲线/路径的像素

Question

是否有解决方案来访问沿曲线 /path 的像素？我们可以使用 LineIterator 来做吗

Answer 1

是的，您可以使用该CvLineIterator方法访问像素。

请参考以下链接，

http://opencv.jp/opencv-2.2_org/c/core_drawing_functions.html

Answer 2

我认为没有任何内置功能。您需要首先在cv::Mat结构中定义线/曲线，然后从那里继续。让我用一个例子来解释。

1. 您有一个图像，cv::Mat input_image并且您使用 acv::HoughLinesDetector来检测图像中存储在cv::Mat hough_lines.
2. 然后，您将需要遍历hough_lines并填充cv::Mat hough_Mat(cv::Size(input_image.size()))（如果您想根据原始数据明亮地显示线条，则应将其转换为 BGR 图像。
3. 然后，简单地遍历hough_Mat哪些像素高于零，然后访问input_image.

尽管此示例是使用霍夫变换的简单示例，但您可以将其与任何其他曲线一起使用，只要您有原始图像中的曲线数据即可。

高温高压

Answer 3

好的，这是一种沿着可以参数化的连接曲线访问像素的方法。可能有更有效的方法，但这个方法非常简单：只需在参数步骤中对曲线进行采样，这样您就不会两次访问一个像素并且不会跳过一个像素：

我以维基百科的参数函数为例： http ://en.wikipedia.org/wiki/Parametric_equation#Some_sophisticated_functions

int main()
{
cv::Mat blank = cv::Mat::zeros(512,512,CV_8U);

// parametric function:
// http://en.wikipedia.org/wiki/Parametric_equation#Some_sophisticated_functions
// k = a/b
// x = (a-b)*cos(t) + b*cos(t((a/b)-1))
// y = (a-b)*sin(t) - b*sin(t((a/b)-1))

float k = 0.5f;
float a = 70.0f;
float b = a/k;

// translate the curve somewhere
float centerX = 256;
float centerY = 256;

// you will check whether the pixel position has moved since the last active pixel, so you have to remember the last one:
int oldpX,oldpY;
// compute the parametric function's value for param t = 0
oldpX = (a-b)*cos(0) + b*cos(0*((a/b)-1.0f)) + centerX -1;
oldpY = (a-b)*sin(0) - b*sin(0*((a/b)-1.0f)) + centerY -1;

// initial stepsize to parametrize the curve
float stepsize = 0.01f;

//counting variables for analyzation
unsigned int nIterations = 0;
unsigned int activePixel = 0;

// iterate over whole parameter region
for(float t = 0; t<4*3.14159265359f; t+= stepsize)
{
    nIterations++;

    // compute the pixel position for that parameter
    int pX = (a-b)*cos(t) + b*cos(t*((a/b)-1.0f)) + centerX;
    int pY = (a-b)*sin(t) - b*sin(t*((a/b)-1.0f)) + centerY;

    // only access pixel if we moved to a new pixel:
    if((pX != oldpX)||(pY != oldpY))
    {
        // if distance to old pixel is too big: stepsize was too big
        if((abs(oldpX-pX)<=1) && (abs(oldpY-pY)<=1))
        {
            //---------------------------------------------------------------
            // here you can access the pixel, it will be accessed only once for that curve position!
            blank.at<unsigned char>((pY),(pX)) = blank.at<unsigned char>((pY),(pX))+1;
            //---------------------------------------------------------------

            // update last position
            oldpX = pX;
            oldpY = pY;

            activePixel++;  // count number of pixel on the contour
        }
        else
        {
            // adjust/decrease stepsize here
            t -= stepsize;
            stepsize /= 2.0f;

            //TODO: choose smarter stepsize updates
        }
    }
    else
    {
        // you could adjust/increase the stepsize here
        stepsize += stepsize/2.0f;

        //TODO: prevent stepsize from becoming 0.0f !!
        //TODO: choose smarter stepsize updates
    }

}
std::cout << "nIterations: " << nIterations << " for activePixel: " << activePixel << std::endl;

cv::imwrite("accessedOnce.png", blank>0);
cv::imwrite("accessedMulti.png", blank>1);

cv::waitKey(-1);
return 0;
}

给出这些结果：

像素访问一次：

像素访问不止一次：

终端输出： nIterations: 1240 for activePixel: 1065