hog适合做行人的识别和车辆识别 对一定区域的形状描述方法

可以表示较大的形状 把图像分成一个一个小的区域的直方图

用cell做单位做直方图 

计算各个像素的梯度强度和方向

用3*3的像素组成一个cell 3*3的cell组成一个block来归一化 提高亮度不变性

常用SVM分类器一起使用 进行行人分类

代码思路：

将图像分成cell为单位 例如把图像分成9*9像素的cell为单位。用sobel计算梯度大小和方向。

遍历每一个cell，一个cell可以分8类，用角度当作数组的下标，也就是分类的依据，数组的大小也就是分类的一个类的大小就是梯度的大小相加。

计算两个图的直方图的直方图距离的大小累加值

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计算hog直方图函数：

int calcHOG(cv::Mat src, float* hist, int nAngle, int cellSize)
{int nX = src.cols / cellSize;int nY = src.rows / cellSize;int binAngle = 360 / nAngle;Mat gx, gy;Mat mag, angle;Sobel(src, gx, CV_32F, 1, 0, 1);Sobel(src, gy, CV_32F, 0, 1, 1);cartToPolar(gx, gy, mag, angle, true);Rect roi;roi.x = 0;roi.y = 0;roi.width = cellSize;roi.height = cellSize;for (int i = 0; i < nY; i++) {for (int j = 0; j < nX; j++) {Mat roiMat;Mat roiMag;Mat roiAgl;roi.x = j * cellSize;roi.y = i * cellSize;//赋值图像roiMat = src(roi);roiMag = mag(roi);roiAgl = angle(roi);//当前cell第一个元素在数组中的位置int head = (i * nX + j) * nAngle;for (int n = 0; n < roiMat.rows; n++) {for (int m = 0; m < roiMat.cols; m++) {//计算角度在哪个bin，通过int自动取整实现int pos = (int)(roiAgl.at<float>(n, m) / binAngle);//以像素点的值为权重hist[head + pos] += roiMag.at<float>(n, m);}}}}return 0;}

mag梯度大小强度  angle是角度的mat

传入的参数就是：图像，直方图数组，分成几个angle类型（一般是8个），cell的大小。

计算两个直方图的距离 

float normL2(float* Hist1, float* Hist2, int size)
{float sum = 0;for (int i = 0; i < size; i++) {sum += (Hist1[i] - Hist2[i]) * (Hist1[i] - Hist2[i]);}sum = sqrt(sum);return sum;
}

第一种是自己申明数组 然后做hog

	Mat temple = imread("hogTemplate.jpg",0);Mat img1 = imread("img1.jpg",0);Mat img2 = imread("img2.jpg",0);float his[3000] = { 0 };float his1[3000] = { 0 };float his2[3000] = { 0 };printf("%d %d\r\n",temple.cols,temple.rows);calcHOG(temple, his, 8, 9);calcHOG(img1, his1, 8, 9);calcHOG(img2, his2, 8, 9);float summ = normL2(his, his1, 3000);float summ2 = normL2(his, his2, 3000);cout << summ <<"\r\n" << endl;cout << "------" << endl;cout << summ2 <<"\r\n" << endl;

用动态开辟内存数组来进行hog

	int nX = refMat.cols / blockSize;int nY = refMat.rows / blockSize;int bins = nX * nY * nAngle;float* ref_hist = new float[bins];memset(ref_hist, 0, sizeof(float) * bins);float* pl_hist = new float[bins];memset(pl_hist, 0, sizeof(float) * bins);float* bg_hist = new float[bins];memset(bg_hist, 0, sizeof(float) * bins);

 这是比较关键的代码 就是动态开辟一个内存

	delete[] ref_hist;delete[] pl_hist;delete[] bg_hist;destroyAllWindows();

记得要释放内存！

完整代码：

	cv::Mat refMat = imread("hogTemplate.jpg");cv::Mat plMat = imread("img1.jpg");cv::Mat bgMat = imread("img2.jpg");int nAngle = 8;int blockSize = 9;int nX = refMat.cols / blockSize;int nY = refMat.rows / blockSize;int bins = nX * nY * nAngle;float* ref_hist = new float[bins];memset(ref_hist, 0, sizeof(float) * bins);float* pl_hist = new float[bins];memset(pl_hist, 0, sizeof(float) * bins);float* bg_hist = new float[bins];memset(bg_hist, 0, sizeof(float) * bins);int reCode = 0;reCode = calcHOG(refMat, ref_hist, nAngle, blockSize);reCode = calcHOG(plMat, pl_hist, nAngle, blockSize);reCode = calcHOG(bgMat, bg_hist, nAngle, blockSize);float dis1 = normL2(ref_hist, pl_hist, bins);float dis2 = normL2(ref_hist, bg_hist, bins);std::cout << "distance between reference and img1:" << dis1 << std::endl;std::cout << "distance between reference and img2:" << dis2 << std::endl;(dis1 <= dis2) ? (std::cout << "img1 is similar" << std::endl) : (std::cout << "img2 is similar" << std::endl);delete[] ref_hist;delete[] pl_hist;delete[] bg_hist;destroyAllWindows();return 0;
}

 

有没有很疑惑 为啥两种计算的方式 他们hog值不一样？

因为第一种我把他灰度化了 所以值偏低，我们现在把第二种方法的也灰度化

 ok 简直一摸一样 结束实验