数字图像处理对比度拉伸直方图

Question

在这里，我附上了我用来绘制对比度图像的直方图并将灰度图像转换为对比度图像的代码。 在这里，我使用低品脱为 122，最高点为 244。在 output 直方图中，它降低了直方图的高度。

我在代码中找不到错误

#include "opencv2/opencv.hpp"
#include "opencv2/highgui.hpp"
#include "opencv2/core.hpp"

using namespace cv;
using namespace std;

int main(int argc, char* argv[]) {
    Mat img = imread(argv[1], 1);

    if (!img.data) { 
        cout << "Could not find the image!" << endl;
        return -1;
    }

    int height = img.rows;
    int width = img.cols;
    int widthstep = img.step;
    int ch = img.channels();

    printf("Height : %d\n", height);
    printf("Width : %d\n", width);
    printf("Widthstep : %d\n", widthstep);
    printf("No of channels : %d\n", ch);

    Mat gray_image(height, width, CV_8UC1, Scalar(0));
    cvtColor(img, gray_image, COLOR_BGR2GRAY);

    Mat new_image = gray_image.clone();
    int v;
    int output{};

    for (int y = 0; y < height; y++) {
        for (int x = 0; x < width; x++) {
            int v = (int)gray_image.at<uchar>(y, x);
            if (v >= 0 && v <= 122) {
                output = int((6 / 122) * v);
            }
            else if (v > 100 && v <= 244) {
                output = int(((244) / (122)) * (v - 122) + 6);
            }
            else if (v > 244 && v <= 255) {
                output = int(((5) / (11)) * (v - 244) + 250);
            }
            new_image.at<uchar>(y, x) = (uchar)output;
        }
    }

    int histn[256];
    for (int i = 0; i < 256; i++) {
        histn[i] = 0;
    }

    for (int y = 0; y < height; y++) {
        for (int x = 0; x < width; x++) {
            histn[(int)new_image.at<uchar>(y, x)] = histn[(int)new_image.at<uchar>(y, x)] + 1;
        }
    }

    for (int i = 0; i < 256; i++) {
        cout << i << ":" << histn[i] << endl;
    }

    int hist_wn = 512;
    int hist_hn = 400;
    int bin_wn = cvRound((double)hist_wn / 256);

    Mat new_histogramImage(hist_hn, hist_wn, CV_8UC1, Scalar(255));

    int maxn = histn[0];
    for (int i = 0; i < 256; i++) {
        if (maxn < histn[i]) {
            maxn = histn[i];
        }
    }

    for (int i = 0; i < 256; i++) {
        histn[i] = ((double)histn[i] / maxn) * new_histogramImage.rows;
    }

    for (int i = 0; i < 256; i++) {
        line(new_histogramImage, Point(bin_wn * (i), hist_hn), Point(bin_wn * (i), hist_hn - histn[i]), Scalar(0), 1, 8, 0);
    }

    imwrite("Gray_Image.png", gray_image);
    imwrite("newcontrast_Image.png", new_image);
    imwrite("Histogram.png", new_histogramImage);

    namedWindow("Image");
    imshow("Image", img);
    namedWindow("Gray_Image");
    imshow("Gray_Image", gray_image);
    namedWindow("newcontrast_Image");
    imshow("newcontrast_Image", new_image);
    namedWindow("New_Histogram");
    imshow("New_Histogram", new_histogramImage);
    namedWindow("Old_Histogram");
    imshow("Old_Histogram", histImage);

    waitKey(0);
    return 0;
}

这是我作为输出得到的新旧直方图

Answer 1

我找到了问题的解决方案。 在这里，我将最低点和最高点值更改为 100 和 240，并在使用这些值时将它们设置为小数。

for (int y = 0; y < height; y++) {
        for (int x = 0; x < width; x++) {
            int v = (int)gray_image.at<uchar>(y, x);
            if (v >= 0 && v <= 100) {
                output = int((5.0/ 100.0) * v);
            }
            else if (v > 100 && v <= 240) {
                output = int(((245.0) / (140.0)) * (v - 100.0) + 5.0);
            }
            else if (v > 240 && v <= 255) {
                output = int(((5.0) / (15.0)) * (v - 240.0) + 250.0);
            }
            new_image.at<uchar>(y, x) = (uchar)output;
        }
    }