opencv学习四

core模块的第二个部分就是OpenCV如何扫描图像、利用查找表和计时：

1、查找表

颜色缩减方法：如果矩阵元素存储的是单通道像素，使用C或C++的无符号字符类型，那么像素可有256个不同值。但若是三通道图像，这种存储格式的颜色数就太多了（确切地说，有一千六百多万种）。用如此之多的颜色可能会对我们的算法性能造成严重影响。其实有时候，仅用这些颜色的一小部分，就足以达到同样效果。所以其做法是：将现有颜色空间值除以某个输入值，以获得较少的颜色数。例如，颜色值0到9可取为新值0，10到19可取为10，以此类推。其公式为：

                                          

即输入的颜色值为：0-9    输出为：0

                 10-19           10

                 20-29           20

                 ……           ……

这样的话，简单的颜色空间缩减算法就可由下面两步组成：一、遍历图像矩阵的每一个像素；二、对像素应用上述公式。

由此可知，对于较大的图像，有效的方法是预先计算所有可能的值，然后需要这些值的时候，利用查找表直接赋值即可。查找表是一维或多维数组，存储了不同输入值所对应的输出值，其优势在于只需读取、无需计算。

计算查找表代码为：

int divideWith; // convert our input string to number - C++ style    stringstream s;    s << argv[2];    s >> divideWith;    if (!s)    {        cout << "Invalid number entered for dividing. " << endl;         return -1;    }        uchar table[256];     for (int i = 0; i < 256; ++i)       table[i] = divideWith* (i/divideWith);

这里我们先使用C++的 stringstream 类，把第三个命令行参数由字符串转换为整数。然后，我们用数组和前面给出的公式计算查找表。

2、计算运行时间

OpenCV提供了两个简便的可用于计时的函数 getTickCount() 和 getTickFrequency() 。第一个函数返回你的CPU自某个事件（如启动电脑）以来走过的时钟周期数，第二个函数返回你的CPU一秒钟所走的时钟周期数。这样，我们就能轻松地以秒为单位对某运算计时：

double t = (double)getTickCount();// 做点什么 ...t = ((double)getTickCount() - t)/getTickFrequency();cout << "Times passed in seconds: " << t << endl;

3、四种方式扫描图像

四种方法的性能如下：

为了得到最优的结果，你最好自己编译并运行这些程序. 为了更好的表现性能差异，我用了一个相当大的图片(2560 X 1600). 性能测试这里用的是彩色图片，结果是数百次测试的平均值.

Efficient Way79.4717 millisecondsIterator83.7201 millisecondsOn-The-Fly RA93.7878 millisecondsLUT function32.5759 milliseconds

我们得出一些结论: 尽量使用 OpenCV 内置函数. 调用LUT 函数可以获得最快的速度. 这是因为OpenCV库可以通过英特尔线程架构启用多线程. 当然,如果你喜欢使用指针的方法来扫描图像，迭代法是一个不错的选择，不过速度上较慢。在debug模式下使用on-the-fly方法扫描全图是一个最浪费资源的方法，在release模式下它的表现和迭代法相差无几，但是从安全性角度来考虑，迭代法是更佳的选择。

#include "stdafx.h"#include <opencv2/core/core.hpp>#include <opencv2/highgui/highgui.hpp>#include <iostream>#include <sstream>using namespace std;using namespace cv;static void help(){cout<< "\n--------------------------------------------------------------------------" << endl<< "This program shows how to scan image objects in OpenCV (cv::Mat). As use case"<< " we take an input image and divide the native color palette (255) with the "  << endl<< "input. Shows C operator[] method, iterators and at function for on-the-fly item address calculation."<< endl<< "Usage:"                                                                       << endl<< "./howToScanImages imageNameToUse divideWith [G]"                              << endl<< "if you add a G parameter the image is processed in gray scale"                << endl<< "--------------------------------------------------------------------------"   << endl<< endl;}Mat& ScanImageAndReduceC(Mat& I, const uchar* table);//通过高效的C风格运算符[]（指针）Mat& ScanImageAndReduceIterator(Mat& I, const uchar* table);//通过安全的迭代法Mat& ScanImageAndReduceRandomAccess(Mat& I, const uchar * table);//通过相关返回值的On-the-fly地址计算int main( int argc, char* argv[]){help();namedWindow("The original picture",1);//命名显示原图像窗口namedWindow("The change picture", 1);//命名显示改变图像窗口//选择显示彩色or灰色char ch=NULL;printf("GRAYSCALE OR COLOR?please enter G/C");scanf("%c",&ch);Mat I, J;if( ch== 'G' )//读取灰色图片I = imread("F:\\1.jpg", CV_LOAD_IMAGE_GRAYSCALE);else//读取彩色图片I = imread("F:\\1.jpg", CV_LOAD_IMAGE_COLOR);if (!I.data){cout << "The image could not be loaded." << endl;return -1;}int divideWith = 0; // 把输入的字符串转为整数stringstream s;s << "50";s >> divideWith;if (!s || !divideWith){cout << "Invalid number entered for dividing. " << endl;return -1;}//根据颜色空间缩减，建立查找表uchar table[256];for (int i = 0; i < 256; ++i)table[i] = (uchar)(divideWith * (i/divideWith));//执行次数为100次const int times = 100;double t;t = (double)getTickCount();for (int i = 0; i < times; ++i){cv::Mat clone_i = I.clone();J = ScanImageAndReduceC(clone_i, table);}//计算100次的平均时间t = 1000*((double)getTickCount() - t)/getTickFrequency();t /= times;//显示原图片和改变图片imshow("The original picture",I);imshow("The change picture",J);waitKey(0);cout << "Time of reducing with the C operator [] (averaged for "<< times << " runs): " << t << " milliseconds."<< endl;t = (double)getTickCount();for (int i = 0; i < times; ++i){cv::Mat clone_i = I.clone();J = ScanImageAndReduceIterator(clone_i, table);}t = 1000*((double)getTickCount() - t)/getTickFrequency();t /= times;cout << "Time of reducing with the iterator (averaged for "<< times << " runs): " << t << " milliseconds."<< endl;t = (double)getTickCount();for (int i = 0; i < times; ++i){cv::Mat clone_i = I.clone();ScanImageAndReduceRandomAccess(clone_i, table);}t = 1000*((double)getTickCount() - t)/getTickFrequency();t /= times;cout << "Time of reducing with the on-the-fly address generation - at function (averaged for "<< times << " runs): " << t << " milliseconds."<< endl;//通过核心函数LUTMat lookUpTable(1, 256, CV_8U);uchar* p = lookUpTable.data;for( int i = 0; i < 256; ++i)p[i] = table[i];t = (double)getTickCount();for (int i = 0; i < times; ++i)LUT(I, lookUpTable, J);t = 1000*((double)getTickCount() - t)/getTickFrequency();t /= times;cout << "Time of reducing with the LUT function (averaged for "<< times << " runs): " << t << " milliseconds."<< endl;char ch1=NULL;scanf("%c",&ch1);while (ch1!='c'){ch1=getchar();}return 0;}Mat& ScanImageAndReduceC(Mat& I, const uchar* const table){// accept only char type matricesCV_Assert(I.depth() != sizeof(uchar));int channels = I.channels();//得到通道数int nRows = I.rows;//获得行数int nCols = I.cols * channels;//行数*通道数=一行中有多少个数值//如果图像矩阵存储空间是连续的if (I.isContinuous()){nCols *= nRows;//得到存储矩阵的大小nRows = 1;}int i,j;uchar* p;for( i = 0; i < nRows; ++i){p = I.ptr<uchar>(i);//得到存储矩阵的起始地址for ( j = 0; j < nCols; ++j){p[j] = table[p[j]];//赋值}}//整个过程起始是将矩阵看出是一个一行nCols列的矩阵，由p指针指向矩阵，再进行赋值。return I;}Mat& ScanImageAndReduceIterator(Mat& I, const uchar* const table){// accept only char type matricesCV_Assert(I.depth() != sizeof(uchar));const int channels = I.channels();switch(channels){case 1:{MatIterator_<uchar> it, end;//定义uchar的迭代器for( it = I.begin<uchar>(), end = I.end<uchar>(); it != end; ++it)*it = table[*it];break;}case 3:{MatIterator_<Vec3b> it, end;//定义Vec3b的迭代器for( it = I.begin<Vec3b>(), end = I.end<Vec3b>(); it != end; ++it){(*it)[0] = table[(*it)[0]];(*it)[1] = table[(*it)[1]];(*it)[2] = table[(*it)[2]];}}}return I;}Mat& ScanImageAndReduceRandomAccess(Mat& I, const uchar* const table){// accept only char type matricesCV_Assert(I.depth() != sizeof(uchar));const int channels = I.channels();switch(channels){case 1:{for( int i = 0; i < I.rows; ++i)for( int j = 0; j < I.cols; ++j )I.at<uchar>(i,j) = table[I.at<uchar>(i,j)];//对行列循环对每个数据点进行赋值break;}case 3:{Mat_<Vec3b> _I = I;//使用Mat_类进行运算，较为方便for( int i = 0; i < I.rows; ++i)for( int j = 0; j < I.cols; ++j ){_I(i,j)[0] = table[_I(i,j)[0]];_I(i,j)[1] = table[_I(i,j)[1]];_I(i,j)[2] = table[_I(i,j)[2]];}I = _I;break;}}return I;}