目的
- 如何遍历图像中的每一个像素？
- OpenCV的矩阵值是如何存储的？
- 如何测试我们所实现算法的性能？
- 查找表是什么？为什么要用它？
测试用例
颜色空间缩减。具体做法就是：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0到9可取为新值0，10到19可取为10。
计算公式：
Lnew = (Lold / 10) * 10
如果对图像矩阵的每一个像素进行这个操作的话，是比较费时的，因为有大量的乘除操作。 这个时候我们的查找表就派上用场了，提前把值计算好，然后要用的时候，直接赋值即可。
- 创建查找表
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);
- 计时
具体用的是getTickCount()和getTickFrequency()两个函数。第一个函数返回的是CPU自某个事件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。
double t = (double)getTickCount();
// 做点什么 ...
t = ((double)getTickCount() - t)/getTickFrequency();
cout << "Times passed in seconds: " << t << endl;
1. 高效的方法 Efficient Way
因为图像中的每个像素是可以顺序存储的，所以可以使用下标进行访问，访问前使用isContinuous()来判断矩阵是否连续存储的。
Mat& ScanImageAndReduceC(Mat& I, const uchar* const table)
{
// accept only char type matrices
CV_Assert(I.depth() != sizeof(uchar));
int channels = I.channels();
int nRows = I.rows * channels;
int nCols = I.cols;
if (I.isContinuous())
{
nCols *= nRows;
nRows = 1;
}
int i,j;
uchar* p;
for( i = 0; i < nRows; ++i)
{
// 获取每一行开始的指针
p = I.ptr<uchar>(i);
for ( j = 0; j < nCols; ++j)
{
p[j] = table[p[j]];
}
}
return I;
}
另外一种方法来实现遍历功能，就是使用data，data会从Mat中返回指向矩阵第一行第一列的指针。注意如果该指针为NULL则表明对象里面无输入，所以这是一种简单的检查图像是否被成功读入的方法。当矩阵是连续存储时，我们就可以通过遍历data来扫描整个图像。
uchar* p = I.data;
for( unsigned int i =0; i < ncol*nrows; ++i)
*p++ = table[*p];
2. 迭代法
Mat& ScanImageAndReduceIterator(Mat& I, const uchar* const table)
{
// accept only char type matrices
CV_Assert(I.depth() != sizeof(uchar));
const int channels = I.channels();
switch(channels)
{
case 1:
{
MatIterator_<uchar> it, end;
for( it = I.begin<uchar>(), end = I.end<uchar>(); it != end; ++it)
*it = table[*it];
break;
}
case 3:
{
MatIterator_<Vec3b> it, end;
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;
}
3. 通过相关返回值的On-the-fly地址计算
Mat& ScanImageAndReduceRandomAccess(Mat& I, const uchar* const table)
{
// accept only char type matrices
CV_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)];
break;
}
case 3:
{
Mat_<Vec3b> _I = I;
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;
}
4. 核心函数LUT (The Core Function)
operationsOnArrays:LUT()
Mat lookUpTable(1, 256, CV_8U);
uchar* p = lookUpTable.data;
for( int i = 0; i < 256; ++i)
p[i] = table[i];
然后我们调用函数(I是输入J是输出)
LUT(I, lookUpTable, J);
性能表现
Efficient Way 79.4717 milliseconds
Iterator 83.7201 milliseconds
On-The-Fly RA 93.7878 milliseconds
LUT function 32.5759 milliseconds
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。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。
四种方法完整的代码：
#include <opencv2/core/core.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <iostream>
#include <sstream>
using namespace std;
using namespace cv;
void help()
{
cout
<< "
--------------------------------------------------------------------------" << 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);
Mat& ScanImageAndReduceIterator(Mat& I, const uchar* table);
Mat& ScanImageAndReduceRandomAccess(Mat& I, const uchar * table);
int main( int argc, char* argv[])
{
help();
if (argc < 3)
{
cout << "Not enough parameters" << endl;
return -1;
}
Mat I, J;
if( argc == 4 && !strcmp(argv[3],"G") )
I = imread(argv[1], CV_LOAD_IMAGE_GRAYSCALE);
else
I = imread(argv[1], CV_LOAD_IMAGE_COLOR);
if (!I.data)
{
cout << "The image" << argv[1] << " could not be loaded." << endl;
return -1;
}
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);
const int times = 100;
double t;
t = (double)getTickCount();
for (int i = 0; i < times; ++i)
J = ScanImageAndReduceC(I.clone(), table);
t = 1000*((double)getTickCount() - t)/getTickFrequency();
t /= times;
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)
J = ScanImageAndReduceIterator(I.clone(), 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)
ScanImageAndReduceRandomAccess(I.clone(), 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;
Mat 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;
return 0;
}
Mat& ScanImageAndReduceC(Mat& I, const uchar* const table)
{
// accept only char type matrices
CV_Assert(I.depth() != sizeof(uchar));
int channels = I.channels();
int nRows = I.rows * channels;
int nCols = I.cols;
if (I.isContinuous())
{
nCols *= nRows;
nRows = 1;
}
int i,j;
uchar* p;
for( i = 0; i < nRows; ++i)
{
p = I.ptr<uchar>(i);
for ( j = 0; j < nCols; ++j)
{
p[j] = table[p[j]];
}
}
return I;
}
Mat& ScanImageAndReduceIterator(Mat& I, const uchar* const table)
{
// accept only char type matrices
CV_Assert(I.depth() != sizeof(uchar));
const int channels = I.channels();
switch(channels)
{
case 1:
{
MatIterator_<uchar> it, end;
for( it = I.begin<uchar>(), end = I.end<uchar>(); it != end; ++it)
*it = table[*it];
break;
}
case 3:
{
MatIterator_<Vec3b> it, end;
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 matrices
CV_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)];
break;
}
case 3:
{
Mat_<Vec3b> _I = I;
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;
}