zoukankan      html  css  js  c++  java
  • OpenCV, color reduction method

    转载请注明出处!!!http://blog.csdn.net/zhonghuan1992


    OpenCV, colorreduction method

    目标:

             这次学习的目标是回答以下的几个问题:

                      1 图片像素是怎样被扫描的?

                       2OpenCV 矩阵值怎样被存储?

                       3怎样衡量算法的性能?

                       4什么是查找表和为什么要用他们?

             看完这篇,希望可以解决上面的这些问题。

    正文:

             首先我们考虑一下简单的色彩减少方法(color reduction method,翻译的不好请指正),假设使用的是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),可是假设使用的是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,所以能够想一些方法来减少这些色彩数量。

             能够使用简单的方法来减少图像色彩空间,比方,将0-9的数字都统一用0来取代,10-19的数字都统一用10取代。这样的转换方案能够用以下的公式表示

             通过上面的公式,把全部像素点的值更新一下。可是,上面的公式中有除法,这里要表达一个是,计算量比較多的情况下,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仅仅有256个,所以我们能够用查找表的方式,我们事先把全部的计算结果都保存在一个数组里,每次要运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。比方32相应30,表table[32]=30是早计算出来的,直接訪问table[32]就OK了。

    图片矩阵怎样在内存中存储的:

    灰度图片的矩阵存储方式:

             灰度图片的每个像素点,仅仅由一个值来表示,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,能够看到,每个像素,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,只是列向量放大了三倍。从图片中能够清楚的看到。

    效率:

             比較像素数量减少方式效率的代码,在本文的最后面,代码看上去非常多,事实上结构比較简单,看一会儿就明确了。附上一张结果图:

            

    最快的OpenCV内的LUT函数。关于LUT,看这里

    能够粗略的看一下代码,代码不难,非常easy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參数,第一个是图像的名字,第二个是參数是公式中的减少颜色数的数字,这里是10,第三个參数,假设是[G]代表是灰度图片,否则不是。
        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);
    
    /*
    	程序主要是看不同的color reduction方式对于程序执行速度的影响。
    	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,来得到执行时间
    
    */
    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来负责将參数转换为数字
        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;
    }


  • 相关阅读:
    机器学习第二次作业
    机器学习上机作业
    机器学习第一次作业
    软工实践个人总结
    第08组 Beta版本演示
    第08组 Beta冲刺(5/5)
    第08组 Beta冲刺(4/5)
    第08组 Beta冲刺(3/5)
    第08组 Beta冲刺(2/5)
    第08组 Beta冲刺(1/5)
  • 原文地址:https://www.cnblogs.com/mfrbuaa/p/3967182.html
Copyright © 2011-2022 走看看