Image capture size is 2000 * 1000 20 20 *

 

digits.png 2000 * 1000, wherein the size of each number is 20 * 20, a total of 5000 ((2000 * 1000) / (20 * 20)) numbers, type of [0-9], [0-9 ] 10 each digit numbers 5000/10 = 500 samples and their sequence into a single image is divided into 20 * 20

#include <the iostream> 
#include <opencv2 \ opencv.hpp>
 the using  namespace CV;
 the using  namespace STD; 

int main () 
{ 
    Mat IMG = imread ( " 1.png " ); 
    Mat Gray; 
    cvtColor (IMG, Gray, CV_BGR2GRAY) ; 
    threshold (Gray, Gray, 0 , 255 , CV_THRESH_BINARY);
     // digits.png is 2000 * 1000, wherein the size of each number is 20 * 20,
     // total of 5000 ((2000 * 1000) / (20 * 20)) numbers, type of [0-9],
     // [0-9] has 10 digits each digital samples 5000/10 = 500
     // into its individual divided images 20 and 20 * serialized (transformed into a one-dimensional array)
    int side = 20;
    int m = gray.rows / side;
    int n = gray.cols / side;
    Mat data, labels;
    for (int i = 0; i < m; i++) 
    {
        int p = 0;
        int offsetRow = i * side;
        for (int j = 0; j < n; j++) 
        {

            int offsetCol = j * side;
            // 截取20*20的小块
            Mat tmp;
            gray(Range(offsetRow, offsetRow + side), Range(offsetCol, offsetCol + side)).copyTo(tmp);
            char path[50];
            char path1[50];
            int num = int(i/5);
            
            sprintf(path, ".\\train\\%d\\", num);
            sprintf(path1, "%d.jpg", p);
            stringstream ss;
            string filename;
            ss << path<< path1;
            ss >> filename;
            imwrite(filename, tmp);
            p++;
        }
    }
    

}