OpenCV, color reduction method
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OpenCV, colorreduction method
目标:
这次学习的目标是回答下面的几个问题:
1 图片像素是如何被扫描的?
2OpenCV 矩阵值如何被存储?
3如何衡量算法的性能?
4什么是查找表和为什么要用他们?
看完这篇,希望能够解决上面的这些问题。
正文:
首先我们考虑一下简单的色彩降低方法(color reduction method,翻译的不好请指正),如果使用的是c或c++无符号的char(八字节大小的空间),一个信道(channel)有256个不同的值(2^8=256),但是如果使用的是GRB方案,三个channel的话,颜色的数量就会变为256*256*256,大概是16个million这么多,这么多的颜色数量,对于计算机来说仍然是一个负担,所以可以想一些方法来降低这些色彩数量。
可以使用简单的方法来降低图像色彩空间,比如,将0-9的数字都统一用0来代替,10-19的数字都统一用10代替。这种转换方案可以用下面的公式表示
通过上面的公式,把所有像素点的值更新一下。但是,上面的公式中有除法,这里要表达一个是,计算量比较多的情况下,不用乘除,就不要用,最好把他们转换为加减。我们知道,在转换前像素点的值只有256个,所以我们可以用查找表的方式,我们事先把所有的计算结果都保存在一个数组里,每次要执行上面的公式计算的时候,结果直接从数组里取出来就ok了。比如32对应30,表table[32]=30是早计算出来的,直接访问table[32]就OK了。
图片矩阵如何在内存中存储的:
灰度图片的矩阵存储方式:
灰度图片的每一个像素点,只由一个值来表示,所以,就是一个普通的二维矩阵。
彩色图片的矩阵存储方式:
彩色图片的存储方式和灰度图片不一样,这里展示的是RGB格式的,可以看到,每一个像素,由三个值,代表蓝色,绿色,红色的三个数值表示,存储方式不是三维的,而是二维,不过列向量放大了三倍。从图片中可以清楚的看到。
效率:
比较像素数量降低方式效率的代码,在本文的最后面,代码看上去很多,其实结构比较简单,看一会儿就明白了。附上一张结果图:
最快的OpenCV内的LUT函数。关于LUT,看这里
可以粗略的看一下代码,代码不难,很容易懂:
#include <opencv2/core/core.hpp>#include <opencv2/highgui/highgui.hpp>#include <iostream>#include <sstream>using namespace std;using namespace cv;static void help(){//这里提示输入有三个参数,第一个是图像的名字,第二个是参数是公式中的降低颜色数的数字,这里是10,第三个参数,如果是[G]代表是灰度图片,否则不是。 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);Mat& ScanImageAndReduceIterator(Mat& I, const uchar* table);Mat& ScanImageAndReduceRandomAccess(Mat& I, const uchar * table);/*程序主要是看不同的color reduction方式对于程序运行速度的影响。使用getTickCount()函数来获取当前时间,利用当前时间-上次获取的时间,来得到运行时间*/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 = 0; // convert our input string to number - C++ style stringstream s; //使用stringstream来负责将参数转换为数字 s << argv[2]; s >> divideWith; if (!s || !divideWith) { cout << "Invalid number entered for dividing. " << endl; return -1; } uchar table[256]; for (int i = 0; i < 256; ++i) table[i] = (uchar)(divideWith * (i/divideWith)); 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); } 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) { 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; 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; int nCols = I.cols * channels; 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;}
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