1、? 图像的遍历
OpenCV图像遍历最高效的方法是指针遍历方法。因为图像在OpenCV里的存储机制问题,行与行之间可能有空白单元(一般是补够4的倍数或8的倍数,有些地方也称作“位对齐”)。这些空白单元对图像来说是没有意思的,只是为了在某些架构上能够更有效率,比如intel MMX可以更有效的处理那种个数是4或8倍数的行。Mat提供了一个检测图像是否连续的函数isContinuous()。当图像连通时,我们就可以把图像完全展开,看成是一行。因此最高效的遍历方法如下:
void imageCopy(const Mat& image,Mat& outImage) {
int nr=image.rows;
int nc=image.cols;
outImage.create(image.size(),image.type());
if(image.isContinuous()&&outImage.isContinuous()) {
nr=1;
nc=nc*image.rows*image.channels();
}
for(int i=0;i<nr;i++) {
const uchar* inData=image.ptr<uchar>(i);
uchar* outData=outImage.ptr<uchar>(i);
for(int j=0;j<nc;j++) {
*outData++=*inData++;
}
}
}
// create the accumulation histograms, img is a binary image, t is 水平或垂直
Mat CalcHistogram(Mat img, int t) {
int sz = (t) ? img.rows : img.cols;
Mat mhist = Mat::zeros(1, sz, CV_32F);
for(int j = 0; j < sz; j++){
Mat data = (t) ? img.row(j) : img.col(j);
// 统计这一行或一列中,非零元素的个数,并保存到mhist中
mhist.at<float>(j) = countNonZero(data);
}
double min, max;
// Normalize histogram
minMaxLoc(mhist, &min, &max);
if (max > 0){
// 用mhist直方图中的最大值,归一化直方图
mhist.convertTo(mhist, -1, 1.0f/max, 0);
}
return mhist;
}
//! 获得字符的特征图
Mat features(Mat in, int sizeData) {
//Histogram features
Mat vhist = ProjectedHistogram(in, VERTICAL);
Mat hhist = ProjectedHistogram(in, HORIZONTAL);
//Low data feature
Mat lowData;
resize(in, lowData, Size(sizeData, sizeData));
//Last 10 is the number of moments components
int numCols = vhist.cols + hhist.cols + lowData.cols * lowData.cols;
Mat out = Mat::zeros(1, numCols, CV_32F);
//Asign values to feature, ANN的样本特征为水平、垂直直方图和低分辨率图像所组成的矢量
int j = 0;
for(int i = 0; i < vhist.cols; i++){
out.at<float>(j) = vhist.at<float>(i);
j++;
}
for(int i = 0; i < hhist.cols; i++){
out.at<float>(j) = hhist.at<float>(i);
j++;
}
for(int x=0; x<lowData.cols; x++){
for(int y=0; y<lowData.rows; y++){
out.at<float>(j)=(float)lowData.at<unsigned char>(x,y);
j++;
}
}
return out;
}
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