使用opencv访问沿曲线/路径的像素 [英] Access to pixels along the curve/path using opencv

问题描述

是否有解决方案来访问沿曲线/路径的像素？我们可以使用LineIterator做它

Is there a solution to access to pixels along the curve /path ? can we use LineIterator to do it

推荐答案

好的，这里是一种访问像素的连接曲线可以参数化的方法。可能有更有效的方法，但这一个很简单：只是在参数步骤中的曲线采样，以便您不访问像素两次，不跳过一个像素：

Ok, here is a way to access pixel along a connected curve that can be parametrized. There might be more efficient ways, but this one is quite simple: just sample the curve in parametersteps so that you don't access a pixel twice and don't skip a pixel:

我使用wikipedia的参数函数作为示例：
http：/ /en.wikipedia.org/wiki/Parametric_equation#Some_sophisticated_functions

I've taken a parametric function from wikipedia as a sample: 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;
}

提供以下结果：

像素已访问一次：

像素被多次访问：

终端输出：
nIterters：1240 for activePixel：1065

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