Question

从（2）均衡图像中我必须创建一个（3）。

原始图片：http://i.imgur.com/X5MKF6z.jpg
均衡图片：http://i.imgur.com/oFBVUJp.png
均衡和拉伸图片：http://i.imgur.com/V7jeaRQ.png

使用OpenCV我可以使用均衡和拉伸的equalizeHist（）。

因此，如果不使用OPENCV，我如何从均衡图像进行拉伸。均衡部分在下面完成。

#include <iostream>
#include <opencv2/highgui/highgui.hpp>
#include <opencv2/imgproc/imgproc.hpp>
#include <opencv/highgui.h>
#include <cstring>
using std::cout;
using std::cin;
using std::endl;

using namespace cv;

void imhist(Mat image, int histogram[])
{

    // initialize all intensity values to 0
    for (int i = 0; i < 256; i++)
    {
        histogram[i] = 0;
    }

    // calculate the no of pixels for each intensity values
    for (int y = 0; y < image.rows; y++)
        for (int x = 0; x < image.cols; x++)
            histogram[(int)image.at<uchar>(y, x)]++;

}

void cumhist(int histogram[], int cumhistogram[])
{
    cumhistogram[0] = histogram[0];

    for (int i = 1; i < 256; i++)
    {
        cumhistogram[i] = histogram[i] + cumhistogram[i - 1];
    }
}

int main()
{
    // Load the image
    Mat image = imread("y1.jpg", CV_LOAD_IMAGE_GRAYSCALE);

    // Generate the histogram
    int histogram[256];
    imhist(image, histogram);


    // Caluculate the size of image
    int size = image.rows * image.cols;
    float alpha = 255.0 / size;

    // Calculate the probability of each intensity
    float PrRk[256];
    for (int i = 0; i < 256; i++)
    {
        PrRk[i] = (double)histogram[i] / size;
    }

    // Generate cumulative frequency histogram
    int cumhistogram[256];
    cumhist(histogram, cumhistogram);

    // Scale the histogram
    int Sk[256];
    for (int i = 0; i < 256; i++)
    {
        Sk[i] = cvRound((double)cumhistogram[i] * alpha);
    }

    // Generate the equlized image
    Mat new_image = image.clone();

    for (int y = 0; y < image.rows; y++)
        for (int x = 0; x < image.cols; x++)
            new_image.at<uchar>(y, x) = saturate_cast<uchar>(Sk[image.at<uchar>(y, x)]);
    //////////////////////////////////////////

    // // Generate the histogram stretched image
    Mat str_image = new_image.clone();

    //for (int a = 0; a < str_image.rows; a++)
    //  for (int b = 0; b < str_image.cols; b++)

    // Display the original Image
    namedWindow("Original Image");
    imshow("Original Image", image);

    // Display equilized image
    namedWindow("Equalized Image");
    imshow("Equalized Image", new_image);


    waitKey();
    return 0;
}

Answer 1

执行此操作的常规方法是找到最暗的像素，并找到最亮的像素。你可以在一个循环循环中迭代你的所有像素，这样的伪代码：

darkest=pixel[0,0]   // assume first pixel is darkest for now, and overwrite later
brightest=pixel[0,0] // assume first pixel is lightest for now, and overwrite later
for all pixels
    if this pixel < darkest
       darkest = this pixel
    else if this pixel > brightest
       brightest = this pixel
    endif
end for

足够简单。所以，让我们说最黑暗和最聪明的人分别是80和220。现在你需要将这个范围80..220拉伸到整个范围0..255。

因此，您从图像中的每个像素中减去80，以在直方图的左端向下移动到零，因此您的范围现在为0..140。所以现在你需要将每个像素乘以255/140以将右端扩展到255.当然，你可以在像素阵列上一次完成两个算术运算。

for all pixels
   newvalue = int((current value - darkest)*255/(brightest-darkest))
end for

C ++拉伸均衡图像

1 个答案: