纹理特征提取

我试图通过使用周围像素的灰度值来提取图像的每个像素的特征向量:http://img59.imageshack.us/img59/7398/texturemap.png标记为黑色的像素是使用的像素，因为其他像素对于稍后使用的SVM结果是冗余的。

当前使用的代码:

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* a指针指向一个包含图像信息的类。我想知道是否有更快的方法来做到这一点。