c++ - 提取纹理特征c++

Question

我试图通过使用周围像素的灰度值来为图像的每个像素提取特征向量: http://img59.imageshack.us/img59/7398/texturemap.png黑色标记的像素是使用的像素，因为其他像素对于后面使用的 SVM 的结果是冗余的。

目前使用这段代码:

vector<Histogram*> texture_based(image_file* image) {
  int cat;
  Mat img = cvLoadImage(image->getName().c_str(), CV_LOAD_IMAGE_GRAYSCALE);
  Mat img_b(img.rows + 12, img.cols + 12, img.depth());

copyMakeBorder(img, img_b, 6, 6, 6, 6, IPL_BORDER_CONSTANT, cvScalarAll(0));

vector<Histogram*> result;

for(int i = 6; i < img_b.rows - 6; ++i) {
    for(int j = 6; j < img_b.cols - 6; ++j) {
        Mat hist = Mat::zeros(1, 49, CV_32FC1);
        cat = 0;
        hist.at<float>(0, 0) = (float)img_b.at<char>(i - 6, j - 6);
        hist.at<float>(0, 1) = (float)img_b.at<char>(i - 5, j - 5);
        hist.at<float>(0, 2) = (float)img_b.at<char>(i - 4, j - 4);
        hist.at<float>(0, 3) = (float)img_b.at<char>(i - 3, j - 3);
        hist.at<float>(0, 4) = (float)img_b.at<char>(i - 2, j - 2);
        hist.at<float>(0, 5) = (float)img_b.at<char>(i - 1, j - 1);
        hist.at<float>(0, 6) = (float)img_b.at<char>(i, j);
        hist.at<float>(0, 7) = (float)img_b.at<char>(i + 1, j + 1);
        hist.at<float>(0, 8) = (float)img_b.at<char>(i + 2, j + 2);
        hist.at<float>(0, 9) = (float)img_b.at<char>(i + 3, j + 3);
        hist.at<float>(0, 10) = (float)img_b.at<char>(i + 4, j + 4);
        hist.at<float>(0, 11) = (float)img_b.at<char>(i + 5, j + 5);
        hist.at<float>(0, 12) = (float)img_b.at<char>(i + 6, j + 6);
        hist.at<float>(0, 13) = (float)img_b.at<char>(i + 6, j - 6);
        hist.at<float>(0, 14) = (float)img_b.at<char>(i + 5, j - 5);
        hist.at<float>(0, 15) = (float)img_b.at<char>(i + 4, j - 4);
        hist.at<float>(0, 16) = (float)img_b.at<char>(i + 3, j - 3);
        hist.at<float>(0, 17) = (float)img_b.at<char>(i + 2, j - 2);
        hist.at<float>(0, 18) = (float)img_b.at<char>(i + 1, j - 1);
        hist.at<float>(0, 19) = (float)img_b.at<char>(i - 1, j + 1);
        hist.at<float>(0, 20) = (float)img_b.at<char>(i - 2, j + 2);
        hist.at<float>(0, 21) = (float)img_b.at<char>(i - 3, j + 3);
        hist.at<float>(0, 22) = (float)img_b.at<char>(i - 4, j + 4);
        hist.at<float>(0, 23) = (float)img_b.at<char>(i - 5, j + 5);
        hist.at<float>(0, 24) = (float)img_b.at<char>(i - 6, j + 6);
        hist.at<float>(0, 25) = (float)img_b.at<char>(i, j - 6);
        hist.at<float>(0, 26) = (float)img_b.at<char>(i, j - 5);
        hist.at<float>(0, 27) = (float)img_b.at<char>(i, j - 4);
        hist.at<float>(0, 28) = (float)img_b.at<char>(i, j - 3);
        hist.at<float>(0, 29) = (float)img_b.at<char>(i, j - 2);
        hist.at<float>(0, 30) = (float)img_b.at<char>(i, j - 1);
        hist.at<float>(0, 31) = (float)img_b.at<char>(i, j + 1);
        hist.at<float>(0, 32) = (float)img_b.at<char>(i, j + 2);
        hist.at<float>(0, 33) = (float)img_b.at<char>(i, j + 3);
        hist.at<float>(0, 34) = (float)img_b.at<char>(i, j + 4);
        hist.at<float>(0, 35) = (float)img_b.at<char>(i, j + 5);
        hist.at<float>(0, 36) = (float)img_b.at<char>(i, j + 6);
        hist.at<float>(0, 37) = (float)img_b.at<char>(i - 6, j);
        hist.at<float>(0, 38) = (float)img_b.at<char>(i - 5, j);
        hist.at<float>(0, 39) = (float)img_b.at<char>(i - 4, j);
        hist.at<float>(0, 40) = (float)img_b.at<char>(i - 3, j);
        hist.at<float>(0, 41) = (float)img_b.at<char>(i - 2, j);
        hist.at<float>(0, 42) = (float)img_b.at<char>(i - 1, j);
        hist.at<float>(0, 43) = (float)img_b.at<char>(i + 1, j);
        hist.at<float>(0, 44) = (float)img_b.at<char>(i + 2, j);
        hist.at<float>(0, 45) = (float)img_b.at<char>(i + 3, j);
        hist.at<float>(0, 46) = (float)img_b.at<char>(i + 4, j);
        hist.at<float>(0, 47) = (float)img_b.at<char>(i + 5, j);
        hist.at<float>(0, 48) = (float)img_b.at<char>(i + 6, j);
        if(image->inAnyRec(i, j))
            cat = 1;

        Mat_<float> new_hist = hist;
        Histogram* t = new Histogram(&new_hist, cat);
        result.push_back(t);
    }
}

return result;
}

其中 image_file* 是指向包含图像信息的类的指针。我想知道是否有更快的方法。

Answer 1

您可以计算 4 遍的操作；每个都将初始化一个包含 12(或 13)个元素的 vector ，向东、向南、东北或东南移动一个像素，并仅替换 vector 中的一个像素。这还需要一次初始化所有直方图 vector (width-12)*(height-12), 49。

支持的选项是将原始图像旋转/倾斜为仅四个数组——您必须分析此时执行 char->float 转换是否有意义。

a b c d   -->  a e i  -->  a f k  >  i f c
e f g h        b f j       b g l     j g d
i j k l        c g k      
               d h l

从这些新数组中，内存读取模式/缓存位置可能会有所不同。