Lab1: Histogram Equalization
1. 实验环境(C++)
- 操作系统版本 MacOS Catalina 10.15
- OpenCV4.0 (imgcodecs | core | highgui | imgproc)
- Cmake-3.14
- Clang-1100.0.33.8
2. 实验步骤
Calculate the histogram H for src
Normalize the histogram.
std::array<double, 256> calNormalizedHist(cv::Mat& source) { std::array<double, 256> acc{0}; // Calculate the histogram H for src for(int i = 0; i < source.rows; i++) for (int j = 0; j < source.cols; j++) acc[ source.ptr<uchar>(i)[j]] ++; // Normalize the histogram. for(int i = 0; i < acc.size(); i++) acc[i] /= source.rows * source.cols; return acc; }
Compute the integral of the histogram: H'
Transform the image using H′ as a look-up table: \[𝚍𝚜𝚝(x,y)=H′(𝚜𝚛𝚌(x,y)) * 255\]
void equalizeHist(cv::Mat& source, cv::Mat& result) { source.copyTo(result); auto hist = calNormalizedHist(source); // Compute the integral of the histogram: H' for(int i = 1; i < 256; i++) hist[i] += hist[i-1]; // Transform the image using H′ as a look-up table: 𝚍𝚜𝚝(x,y)=H′(𝚜𝚛𝚌(x,y)) * 255 for(int i = 0; i < source.rows; i++) for (int j = 0; j < source.cols; j++) result.ptr<uchar>(i)[j] = hist[source.ptr<uchar>(i)[j]] * 255; }
Test Code:
cv::Mat cv, lab, m = cv::imread("../data/lena.png", cv::IMREAD_GRAYSCALE); cv::equalizeHist(m, cv); equalizeHist(m, lab); cv::imwrite("../out/m.png", m); cv::imwrite("../out/cv.png", cv); cv::imwrite("../out/lab.png", lab); cv::imwrite("../out/cv-m.png", abs(cv-m)); cv::imwrite("../out/lab-m.png", abs(lab-m)); cv::imwrite("../out/lab-cv.png", abs(lab-cv)); cv::imwrite("../out/hist.m.png", drawHist(calNormalizedHist(m))); cv::imwrite("../out/hist.cv.png", drawHist(calNormalizedHist(cv))); cv::imwrite("../out/hist.lab.png", drawHist(calNormalizedHist(lab))); cv::imwrite("../out/acchist.m.png", drawHist(calNormalizedHist(m), true)); cv::imwrite("../out/acchist.cv.png", drawHist(calNormalizedHist(cv), true)); cv::imwrite("../out/acchist.lab.png", drawHist(calNormalizedHist(lab),true));
3. 实验结果

image: Origin

image: OpenCV

image: Result

dif: Origin&OpenCV

dif: Result&Origin

dif: Result&OpenCV

Hist: Origin

Hist: OpenCV

Hist: Result

AccHist: Origin

AccHist: OpenCV

AccHist: Result
4. 实验结果分析
明显本文实现的算法与OpenCV实现的高度一致(不考虑指令集优化: SIMD|SEE4 etc.)
由Histogram与Accumulate Histogram上看, 连续变量的均匀分布意味着概率分布函数满足线性分布, 与理论推导得出的性质一致
从直方图直观观察和算法分析, 易知基于直接灰度映射(点对点映射)的经典直方图均衡算法会导致灰阶的减少.因而加剧色带(Banding)现象:
- Lookup-Table \(H’\)在定义上不是单射
所以这个意义上, 直方图的均衡是一个可以把相邻柱子合并的挪动柱子的游戏~
5 Attachment
本部分是附加部分, 主要回答实验课程上与老师的一个小小的 argument:
怎么得到均衡后的直方图 ?
比较常见的方法: 图像均衡后再统计一次直方图
比较少用的方法: 直接均衡原直方图
std::array<double, 256> equalizeHist(std::array<double, 256>&& hist) { std::array<double, 256> hist_eq{0}, hist_acc{0}; hist_acc[0] = hist[0]; // Compute the integral of the histogram: H' for(int i = 1; i < 256; i++) hist_acc[i] = hist_acc[i-1] + hist[i]; for(int i = 0; i < 256; i++) hist_eq[hist_acc[i]*255] += hist[i]; return hist_eq; }
先计算均衡后的直方图再均衡图像对吗?
- 不对, 均衡是一种方法, 可以直接均衡图像然后计算新的直方图, 也可以直接均衡直方图, 两直方图一致
直方图均衡的均衡指什么?
均衡一种概率变换, 可以将线性的目标概率分布函数拓展到任意的分布函数:
Transformation between two distribution funtion.5.* 下面给出指定累计直方图的均衡结果的部分例子
- \[y=x^2\]
AccHist
Image
Hist:
Hist: Directly - \[y=x^5\]
AccHist
Image
Hist:
Hist: Directly - \[y=sin(x)\]
AccHist
Image
Hist:
Hist: Directly - \[y=\sqrt{2x-x^2}\]
AccHist
Image
Hist:
Hist: Directly \[y=e^{x-1}\]
AccHist
Image
Hist:
Hist: Directly