提取纹理特征C++
问题描述：
我试图通过使用周围像素的灰度值为图像的每个像素提取特征向量： 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的结果是多余的。提取纹理特征C++
目前，该代码使用：
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;
}
凡IMAGE_FILE *指针是有关图像信息的类。 我想知道是否有更快的方法做到这一点。
答
您可以计算4遍的操作;每个将初始化12（或13）个元素的向量，向东，南，东北或东南移动一个像素，并从该向量中仅替换一个像素。这也需要初始化所有的直方图向量（width-12）*（height-12），49。
支持选项是将原始图像旋转/倾斜到四个数组 - 如果在该点执行char-> float转换有意义，则必须进行配置。
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/缓存位置可以有所不同。
也许尝试cv :: filter2D（）或cv :: bitwise_and（）代码中的某处。 – William 2013-04-24 10:21:24