项目功能实现：对一张图像进行统计最大、最小像素值、均差以及方差的值
按照之前的博文结构来，这里就不在赘述了

一、头文件

pixel_statistic.h

#pragma once

#include<opencv2/opencv.hpp>

using namespace cv;

class Pixel_Statistic {
public:
	void pixel_statistic(Mat& image);
};

#pragma once

二、函数实现

pixel_statistic.cpp

double min_value, max_value;
因为像素值是浮点数，故接收的数据类型定义为double

std::vector<Mat>yy;
split(image, yy);
对三颜色通道的图片进行通道分割
具体分割实现思路可参考博文：十一、图像颜色通道分离与合并

minMaxLoc(yy[i], &min_value, &max_value, &min_location, &max_location, Mat());
需要循环遍历每个通道，并统计计算最大、最小像素值的数值和位置
OpenCV颜色通道为BGR，yy[0]是B通道

meanStdDev(image, mean, stddev);
得到的是三颜色通道的均值和方差总和，不是单通道的均值和方差
获取每个颜色通道的均值和方差，对yy进行统计
yy[0]是B通道，BGR以此类推即可

#include"pixel_statistic.h"
#include<iostream>
#include<opencv2/opencv.hpp>

void Pixel_Statistic::pixel_statistic(Mat& image) {
	double min_value, max_value;
	Point min_location, max_location;
	Mat mean, stddev;

	std::vector<Mat>yy;
	split(image, yy);//minMaxLoc统计的是单通道的图片，故需要通过split将三通道进行分离
	for (int i = 0; i < yy.size(); i++) {//每个不同颜色通道依次进行统计
		minMaxLoc(yy[i], &min_value, &max_value, &min_location, &max_location, Mat());
		std::cout << "NO.channels is: " << i << " min_value is: " << min_value <<  " min_location is:" << min_location << " max_value is : " << max_value <<" max_location is : "<<max_location << std::endl;
	}

	meanStdDev(image, mean, stddev);//得到的是三颜色通道的均值和方差总和，不是单通道的均值和方差
	std::cout << "all channel's means is; \n" << mean << std::endl;
	std::cout << "all channel's stddev is; \n" << stddev << std::endl;
	//想要获得单通道的均值和方差，我的想法思路是 通过split进行分离通道，然后挨个进行计算统计即可
}

三、主函数

yy_main.cpp

#include <opencv2/opencv.hpp>
#include <iostream>
#include "pixel_statistic.h"

using namespace cv;
using namespace std;

int main(int argc, char** argv) {
	Mat src = cv::imread("E:/C++_workspace/beyond.jpg", IMREAD_COLOR);

	if (src.empty()) {
		printf("load image is false...\n");
		return -1;
	}

	namedWindow("yanyu", WINDOW_FREERATIO);
	imshow("yanyu", src);

	Pixel_Statistic yy;
	yy.pixel_statistic(src);

	waitKey(0);
	destroyAllWindows();

	return 0;
}

项目结构如下：

运行结果如下：