与已知图像匹配的局部二值模式 我正在寻找一种使用OpenCV和C++实现局部二进制模式的方法。

目前我发现：

然而，我需要比较两幅图像或LBP直方图，并给出一些相似性指数

这是我修改过的代码：

    #include <opencv/cv.h>
#include <opencv/highgui.h>
#include "lbp.hpp"
#include "histogram.hpp"

using namespace cv;

int main(int argc, const char *argv[]) {
    int deviceId = 0;
    if(argc > 1)
        deviceId = atoi(argv[1]);

    VideoCapture cap(deviceId);

    if(!cap.isOpened()) {
        cerr << "Capture Device ID " << deviceId << "cannot be opened." << endl;
        return -1;
    }

    // initial values
    int radius = 1;
    int neighbors = 8;

    // windows
    namedWindow("original",CV_WINDOW_AUTOSIZE);
    namedWindow("lbp",CV_WINDOW_AUTOSIZE);

    // matrices used
    Mat test;
    Mat test1;
    Mat frame; // always references the last frame
    Mat dst; // image after preprocessing
    Mat dst1;
    Mat lbp; // lbp image
    Mat lbp1;

    // just to switch between possible lbp operators
    vector<string> lbp_names;
    lbp_names.push_back("Extended LBP"); // 0
    lbp_names.push_back("Fixed Sampling LBP"); // 1
    lbp_names.push_back("Variance-based LBP"); // 2
    int lbp_operator=1;

    bool running=true;
    while(running) {
        //cap >> frame;
        dst = imread("Coin1.jpg", CV_LOAD_IMAGE_GRAYSCALE); //Known Image
        dst1 = imread("Coin2.jpg", CV_LOAD_IMAGE_GRAYSCALE); //Compared to

        switch(lbp_operator) {
        case 0:
            lbp::ELBP(test, lbp, radius, neighbors); // use the extended operator
            break;
        case 1:
            lbp::OLBP(dst, lbp); // use the original operator
            lbp::OLBP(dst1, lbp1); // use the original operator
            break;
        case 2:
            lbp::VARLBP(dst, lbp, radius, neighbors);
            break;
        }
        // now to show the patterns a normalization is necessary
        // a simple min-max norm will do the job...
        normalize(lbp, lbp, 0, 255, NORM_MINMAX, CV_8UC1);

        Mat lbp_hist, lbp1_hist;
        int histSize[] = {256};
        float s_ranges[] = { 0, 256 };
        const float* ranges[] = { s_ranges };

    // Use the o-th and 1-st channels
    int channels[] = { 0 };

        calcHist( &lbp, 1, channels, Mat(), lbp_hist, 1, histSize, ranges, true, false );
        normalize( lbp1_hist, lbp1_hist, 0, 1, NORM_MINMAX, -1, Mat() );

        calcHist( &lbp1, 1, channels, Mat(), lbp1_hist, 1, histSize, ranges, true, false );
        normalize( lbp_hist, lbp_hist, 0, 1, NORM_MINMAX, -1, Mat() );

        double base_base = compareHist( lbp_hist, lbp1_hist, 0 );
        printf("%f\n",base_base); //get a similarity

        //imshow("original", lbp);
        //imshow("lbp", lbp1);
        imshow("1", lbp_hist);
        imshow("2", lbp1_hist);

        char key = (char) waitKey(0);;

    }
        return 0; // success
}

#包括
#包括
#包括“lbp.hpp”
#包括“直方图.hpp”
使用名称空间cv；
int main（int argc，const char*argv[]{
int deviceId=0；
如果（argc>1）
deviceId=atoi（argv[1]）；
视频捕获帽（deviceId）；
如果（！cap.isOpened（））{
cerr我记得在开始使用OpenCV LBPH时遇到了类似的问题

尝试使用此函数获取直方图

void lbp::histogram(const Mat& src, Mat& hist, int numPatterns) {
switch(src.type()) {
    case CV_8SC1: histogram_<char>(src, hist, numPatterns); break;
    case CV_8UC1: histogram_<unsigned char>(src, hist, numPatterns); break;
    case CV_16SC1: histogram_<short int>(src, hist, numPatterns); break;
    case CV_16UC1: histogram_<unsigned short>(src, hist, numPatterns); break;
    case CV_32SC1: histogram_<int>(src, hist, numPatterns); break;
    }
}


template <typename _Tp>
void lbp::histogram_(const Mat& src, Mat& hist, int numPatterns) {
    hist = Mat::zeros(1, numPatterns, CV_32SC1);
    for(int i = 0; i < src.rows; i++) {
        for(int j = 0; j < src.cols; j++) {
            int bin = src.at<_Tp>(i,j);
            hist.at<int>(0,bin) += 1;
        }
    }
 }

//Manual normalization 
cv::Mat hist_norm=cv::Mat::zeros(1,hist.cols,CV_32F);
int sum=0;
for(int j=0;j<hist.cols;j++){sum+=hist.at<int>(0,j);}
for(int j=0;j<hist.cols;j++){hist_norm.at<float>(0,j)+= (float)hist.at<int>(0,j)/(float)sum;}

void lbp:：直方图（常量Mat和src、Mat和hist、int numPatterns）{
开关（src.type（））{
案例CV_8SC1：直方图（src、hist、numPatterns）；中断；
案例CV_8UC1：直方图（src、hist、numPatterns）；中断；
案例CV_16SC1：直方图（src、hist、numPatterns）；中断；
案例CV_16UC1：直方图（src、hist、numPatterns）；中断；
案例CV_32SC1：直方图（src、hist、numPatterns）；中断；
}
}
模板
无效lbp：：直方图（常数矩阵和src、矩阵和历史、整数）{
hist=Mat:：零（1，numPatterns，CV_32SC1）；
对于（int i=0；i对于（int j=0；jI厌倦了用lbp:：histogram（lbp，lbp_hist，2）来称呼它）；但是最终得到了一个类似的微小的histogram。你是说直方图的大小像256x1吗？这是正常的。