8.直方图

一.直方图

简单的说灰度直方图就是：一幅图像中灰度级与出现这种灰度的概率之间关系的图形，灰度直方图是一种统计表达，反映了不同灰度级出现的统计概率（个数），其横坐标是：灰度级，纵坐标是：出现的个数（概率）。

（1）直方图的离散函数 h(rk)=nk。其中rk是第K级的灰度值，nk是图像中灰度为rk的像素的个数。

（2）归一化直方图 p(rk)=nk/MN。其中k = 0,1,2.....L-1（灰度级的范围是[ 0 , L-1]），MN表示像素的总的个数。

算法实现：

 

 

Mat histogram_normalization(Mat& img, int a, int b)
{
    int width = img.cols;
    int height = img.rows;
    int channel = img.channels();

    int c = 0, d = 0; //图片灰度阈值
    int val;
    Mat out = Mat::zeros(height, width, CV_8UC3);

    //get c, d
    for (int y = 0; y < height; y++)
    {
        for (int x = 0; x < width; x++)
        {
            for (int _c = 0; _c < channel; _c++)
            {
                val = (float)img.at(y, x)[_c];
                c = fmin(val, c);
                d = fmax(val, d);
            }
        }
    }

    // histogram transformation
    for (int y = 0; y < height; y++)
    {
        for (int x = 0; x < width; x++)
        {
            for (int _c = 0; _c < 3; _c++)
            {
                val = (float)img.at(y, x)[_c];
                if (val < a)
                    out.at(y, x)[_c] = (uchar)a;
                else if (val >= c && val < d)
                    out.at(y, x)[_c] = (uchar)(b - a) / (d - c)*(val - c) + a;
                else
                    out.at(y, x)[_c] = (uchar)b;
            }
        }
    }
    return out;
}

二. 直方图操作

 

 

 这种操作不能将图像灰度等级分布在整个灰度区域，但可以提高对比度。

 算法实现：

 1 cv::Mat histogram_transform(cv::Mat img, int m0, int s0)
 2 {
 3     // get height and width
 4     int width = img.cols;
 5     int height = img.rows;
 6     int channel = img.channels();
 7     cv::Mat out = cv::Mat::zeros(height, width, CV_8UC3);
 8     double val = 0;
 9     double sum = 0.0, squared_sum = 0.0;
10     for (int y = 0; y < height; y++)
11     {
12         for (int x = 0; x < width; x++)
13         {
14             for (int c = 0; c < channel; c++)
15             {
16                 val = (float)img.at(y, x)[c];
17                 sum += val;
18                 squared_sum += (val * val);
19             }
20         }
21     }
22     double m, sigma;
23     //期望
24     m = sum / (height*width*channel);
25     //标准差
26     sigma = sqrt(squared_sum / (height*width*channel) - m * m);
27 
28     cout << "期望 ： " << m << endl;
29     cout<<"标准差"<< sigma<<endl;
30     // histogram transformation
31     for (int y = 0; y < height; y++)
32     {
33         for (int x = 0; x < width; x++)
34         {
35             for (int c = 0; c < channel; c++)
36             {
37                 val = img.at(y, x)[c];
38                 out.at(y, x)[c] = (uchar)(s0 / sigma * (val - m) + m0);
39             }
40 
41         }
42     }
43     return out;
44 }

三、均衡化

 

 

 算法实现：

 1 // histogram equalization
 2 cv::Mat histogram_equalization(cv::Mat img)
 3 {
 4     // get height and width
 5     int width = img.cols;
 6     int height = img.rows;
 7     int channel = img.channels();
 8 
 9     // output image
10     cv::Mat out = cv::Mat::zeros(height, width, CV_8UC3);
11 
12     // histogram equalization hyper-parameters
13     double Zmax = 255;
14     double hist[255] = { 0 };
15     double S = height * width*channel;
16     int val = 0;
17     //统计像素个数
18     for (int y = 0; y < height; y++)
19     {
20         for (int x = 0; x < width; x++)
21         {
22             for (int c = 0; c < channel; c++)
23             {
24                 val = (int)img.at(y, x)[c];
25                 hist[val]++;
26             }
27         }
28     }
29 
30     double hist_sum = 0;
31     //均衡化处理
32     for (int y = 0; y < height; y++)
33     {
34         for (int x = 0; x < width; x++)
35         {
36             for (int c = 0; c < channel; c++)
37             {
38                 val = (int)img.at(y, x)[c];
39                 hist_sum = 0;
40                 for (int k = 0; k < val; k++)
41                 {
42                     hist_sum += hist[k];
43                 }
44                 out.at(y, x)[c] = (uchar)(Zmax / S * hist_sum);
45             }
46         }
47     }
48     

 

 四、算法推到

   https://blog.csdn.net/u013355826/article/details/57417771