Question

更新我问的问题。我已经成功地从RGB转换为HSI并再次返回。在我输出图像之前，我的教授给我指示通过在I（强度）上添加（（1.0 - I）/ 2.0）来增加图像强度。当我这样做时，它似乎导致溢出，因此我所有的检查RGB等。如果我减去上面的公式，它成功地使图像变暗而没有溢出问题。我似乎无法确定这种溢出的原因，也不知道如何纠正它。我起初认为它与R = G = B的值有关，但我通过输出这些变量并调试确认情况并非如此。造成这种情况的原因是饱和度S等于零。当S = 0时，我知道色调H是无关紧要的。任何帮助将不胜感激！

#include<opencv2/core/core.hpp>
#include<opencv2/highgui/highgui.hpp>
#include<opencv2/imgproc/imgproc.hpp>

#include<iostream>

#define M_PI (3.14159265358979323846)

using namespace cv;


int main(int argc, char * argv[])
{
if (argc <= 1)
{
    std::cerr << "Not enough arguments." << std::endl;
    return (1);
}

Mat img = cv::imread(argv[1], CV_LOAD_IMAGE_COLOR);
std::cout << img.channels() << std::endl;

if (img.empty())
{
    std::cerr << "Unable to open the image " << argv[1] << "." << std::endl;
    return (1);
}

namedWindow("Original Image", CV_WINDOW_AUTOSIZE);
imshow("Original Image", img);

Mat hsi(img.rows, img.cols, img.type());
Mat newBGRimg = Mat::zeros(img.rows, img.cols, img.type());

double B, G, R, H, S, I, b, g, re;
int intB = 0, intG = 0, intR = 0;

for (int r = 0; r < img.rows; r++)
{
    for (int c = 0; c < img.cols; c++)
    {
        b = img.at<Vec3b>(r, c)[0];
        g = img.at<Vec3b>(r, c)[1];
        re = img.at<Vec3b>(r, c)[2];

        // Normalize colors
        B = b / 255;
        G = g / 255;
        R = re / 255;

        // BGR to HSI           
        double theta = acos((((R - G) + (R - B)) / 2) / (sqrt((pow((R - G), 2)) + ((R - B)*(G - B) + 0.0001))));

        if (B <= G)
        {
            H = theta;
        }

        if (B > G)
        {
            H = (2 * M_PI) - theta;
        }

        if (H < 0)
            H = 0;
        if (H > 2 * M_PI)
            H = 2 * M_PI;

        double min = std::min({ B, G, R });
        S = 1 - (3 * min) / (B + G + R);    // / (B + G + R)
        if (S < 0)
            S = 0;
        if (S > 1)
            S = 1;

        I = (B + G + R) / 3;

        I = I + ((1.0 - I) / 2.0); 
        if (I < 0)
            I = 0;
        if (I > 1)
            I = 1;

       // HSI to BGR
        if (H >= 0 && H < (2 * M_PI / 3))
        {

            B = I * (1 - S);
            R = I * (1 + ((S * cos(H)) / (cos((M_PI / 3) - H))));
            G = (3 * I) - (R + B);

            B *= 255;
            G *= 255;
            R *= 255;

            if (B > 255)
                B = 255;
            if (G > 255)
                G = 255;
            if (R > 255)
                R = 255;
            if (B < 0)
                B = 0;
            if (G < 0)
                G = 0;
            if (R < 0)
                R = 0;

            intB = (int)round(B);
            intG = (int)round(G);
            intR = (int)round(R);
            //std::cout << intB << " " << intG << " " << intG <<  " " << r << " " << c << std::endl;

            newBGRimg.at<Vec3b>(r, c)[0] = intB;
            newBGRimg.at<Vec3b>(r, c)[1] = intG;
            newBGRimg.at<Vec3b>(r, c)[2] = intR;
        }
        else if (H >= (2 * M_PI / 3) && H < (4 * M_PI / 3))
        {
            H = H - (2 * M_PI) / 3;
            R = I * (1 - S);
            G = I * (1 + ((S * cos(H)) / (cos((M_PI / 3) - H))));
            B = (3 * I) - (R + G);

            B *= 255;
            G *= 255;
            R *= 255;

            if (B > 255)
                B = 255;
            if (G > 255)
                G = 255;
            if (R > 255)
                R = 255;
            if (B < 0)
                B = 0;
            if (G < 0)
                G = 0;
            if (R < 0)
                R = 0;

            intB = (int)round(B);
            intG = (int)round(G);
            intR = (int)round(R);

            newBGRimg.at<Vec3b>(r, c)[0] = intB;
            newBGRimg.at<Vec3b>(r, c)[1] = intG;
            newBGRimg.at<Vec3b>(r, c)[2] = intR;
        }
        else if (H >= (4 * M_PI / 3))
        {
            H = H - (4 * M_PI) / 3;
            G = I * (1 - S);
            B = I * (1 + ((S * cos(H)) / (cos((M_PI / 3 - H)))));
            R = (3 * I) - (G + B);

            B *= 255;
            G *= 255;
            R *= 255;

            if (B > 255)
                B = 255;
            if (G > 255)
                G = 255;
            if (R > 255)
                R = 255;
            if (B < 0)
                B = 0;
            if (G < 0)
                G = 0;
            if (R < 0)
                R = 0;

            intB = (int)round(B);
            intG = (int)round(G);
            intR = (int)round(R);

            newBGRimg.at<Vec3b>(r, c)[0] = intB;
            newBGRimg.at<Vec3b>(r, c)[1] = intG;
            newBGRimg.at<Vec3b>(r, c)[2] = intR;
        }
    }
}
namedWindow("New RGB Image", CV_WINDOW_AUTOSIZE);
imshow("New RGB Image", newBGRimg);
imwrite("hsi-to-rgb.jpg", newBGRimg);

waitKey(0);
return 0;
}

Answer 1

在我的原始代码中，我试图使用ex来规范化BGR值。 B / B + G + R每DI第3版补充材料。。这应该是B / 255以将其标准化为[0,1]。我相信这本书在建议这种规范化方法时是不正确的。

使用C ++在OpenCV中将HSI转换为RGB时增加强度

1 个答案: