首先需要说明的是,这个系统是我们大二下学期的二级项目。
正因为是二级项目,所以老师要求我们不能使用现成的库(如 zbar)和现有的算法(如 KNN 算法)。
所幸,老师给的图片也并不复杂,类似下图:
我们需要做的工作便是找到并截取红框区域,将字符分割然后识别。
大体思路:
- 倾斜图像修正
- 截取 ISBN 号所在区域
- 字符分割
- 字符识别
1. 倾斜图像修正
- 截取 ISBN 号所在行
- 字符分割
- 字符识别
详细代码:
#include<opencv.hpp> #include<iostream> #include<vector> #include<string>
using namespace cv; using namespace std; //计算修正角度
double GetTurnTheta(Mat inputImg) { //计算垂直方向导数
Mat yImg; Sobel(inputImg, yImg, -1, 0, 1, 5); //直线检测
vector<Vec2f>lines; HoughLines(yImg, lines, 1, CV_PI / 180, 180); //计算旋转角度
float thetas = 0; for (int i = 0; i < lines.size(); i++) { float theta = lines[i][1]; thetas += theta;
}
if (lines.size() == 0) {//未检测到直线
thetas = CV_PI / 2; } else {//检测到直线,取平均值
thetas /= lines.size(); } return thetas; } //寻找 ISBN 所在行
void FindRowRanges(Mat inputImg, int thresh, int mnRow, int mxRow, int mnsize, int &st, int &ed) { //边缘检测,方便找到梯度大的地方,忽略梯度小的地方
Mat cannyNums; blur(inputImg, cannyNums, Size(3, 3)); Canny(cannyNums, cannyNums, thresh, thresh * 2, 3); //寻找上下边界
for (int i = mnRow; i < mxRow; i++) { if (cannyNums.at<uchar>(i, 0) != 0) { st = i; break; } } for (int i = mxRow; i >= mnRow; i--) { if (cannyNums.at<uchar>(i, 0) != 0) { ed = i; break; } } //范围过小,调整二值化阈值,重新寻找
if (abs(ed - st) < mnsize) { thresh -= 10; if (thresh <= 0) { st = mnRow; ed = mxRow; return; } FindRowRanges(inputImg, thresh, mnRow, mxRow, mnsize, st, ed); } } //寻找每个字符对应位置
void FindColRanges(Mat inputImg,vector<float>&pts) { int thre = 0; for (int j = 1; j < inputImg.cols - 1; j++) { if (inputImg.at<uchar>(0, j) > thre && inputImg.at<uchar>(0, j - 1) <= thre) {//左边缘
pts.push_back(j - 1); } else if (inputImg.at<uchar>(0, j) > thre && inputImg.at<uchar>(0, j + 1) <= thre){//右边缘
pts.push_back(j + 1); } } } //模板匹配
bool Comp(pair<int, int>a, pair<int, int>b) { return a.second < b.second; } int CalcImg(Mat inputImg) { int nums = 0; for (int i = 0; i < inputImg.rows; i++) { for (int j = 0; j < inputImg.cols; j++) { if (inputImg.at<uchar>(i, j) != 0) { nums += inputImg.at<uchar>(i, j); } } } return nums; } //模板匹配的主要函数
char CheckImg(Mat inputImg, int k) { //读取模板图片
string sampleImgPath = "样例/*.jpg"; vector<String> sampleImgFN;
glob(sampleImgPath, sampleImgFN, false); int sampleImgNums = sampleImgFN.size(); pair<int, int>*nums = new pair<int, int>[sampleImgNums];//first 记录模板的索引号,second 记录两图像之差
for (int i = 0; i < sampleImgNums; i++) { nums[i].first = i; Mat numImg = imread(sampleImgFN[i], 0); Mat delImg; absdiff(numImg, inputImg, delImg); nums[i].second = CalcImg(delImg); } sort(nums, nums + sampleImgNums, Comp);//选择差值最小的模板
int index = nums[0].first / 2; switch (index) { case 0: case 1: case 2: case 3: case 4: case 5: case 6: case 7: case 8: case 9: return index + '0'; case 10: return 'I'; case 11: return 'S'; case 12: return 'B'; case 13: return 'N'; case 14: return 'X'; default: return ' '; } } int main() { int rtNums = 0, accNums = 0, sunNums = 0;//分别代表:正确的数量,被准确识别的字符的数量,要识别的字符的总和 //读取 ISBN 图片
string testImgPath = "数据集/*.jpg"; vector<String> testImgFN;//必须cv的String
glob(testImgPath, testImgFN, false); int testImgNums = testImgFN.size(); for (int index =0; index < testImgNums; index++) { //int index = 25; //调整原图大小
Mat src = imread(testImgFN[index]); double width = 400; double height = width * src.rows / src.cols; resize(src, src, Size(width, height)); //转换成二值图像
Mat binImg; cvtColor(src, binImg, COLOR_BGR2GRAY); threshold(binImg, binImg, 0, 255, THRESH_BINARY_INV | THRESH_OTSU); //计算调整角度
double thetas = GetTurnTheta(binImg); thetas = 180 * thetas / CV_PI - 90; //旋转二值图像
Mat turnBin; Mat M = getRotationMatrix2D(Point(width / 2, height / 2), thetas, 1); warpAffine(binImg, turnBin, M, src.size()); //计算每行点数
Mat rowNums = Mat(src.rows, 1, CV_8UC1); int st = - 1, ed = - 1;//起始和终止行
for (int i = 0; i < src.rows; i++) { int temC = 0; for (int j = 0; j < src.cols; j++) {//统计每行像素点个数
if (turnBin.at<uchar>(i, j) != 0) { temC++; } } rowNums.at<uchar>(i, 0) = temC; } //寻找截取范围,并适当扩大截取范围
FindRowRanges(rowNums, 110, 0, src.rows / 4, 10, st, ed); int adds = 4; st = st >= adds ? (st -= adds) : 0; ed -= adds; //弥补旋转缺失区域
Mat background = Mat(src.rows, src.cols, CV_8UC1, Scalar(255)); warpAffine(background, background, M, src.size()); bitwise_not(background, background);
Mat turnSrc; warpAffine(src, turnSrc, M, src.size()); src.copyTo(turnSrc, background); //截取 ISBN 所在行
Mat subImg = Mat(turnSrc, Range(st, ed), Range(0, turnSrc.cols));//截取原图相应部分 //调整大小
width = 900; height = width * subImg.rows / subImg.cols; resize(subImg, subImg, Size(width, height)); //转换为二值图像
binImg = Mat(); cvtColor(subImg, binImg, COLOR_BGR2GRAY); threshold(binImg, binImg, 0, 255, THRESH_BINARY_INV | THRESH_OTSU);//计算每列点数
Mat colNums = Mat::zeros(1, subImg.cols, CV_8UC1); for (int i = 0; i < subImg.rows; i++) { for (int j = 1; j < subImg.cols - 1; j++) {//统计每行像素点个数
if (binImg.at<uchar>(i, j) != 0) { colNums.at<uchar>(0, j)++; } } }//寻找字符边界
vector<float>pts; FindColRanges(colNums, pts); //截取字符并识别
string result = ""; for (int j = 0; j < pts.size(); j += 2) {//j 为左边界,j+1 为右边界//截取当前字符所在区域,方便后续操作
Mat roi = Mat(binImg, Range(0, subImg.rows), Range(pts[j], pts[j + 1])); Mat roiImg; roi.copyTo(roiImg); //寻找最小正矩形,并排除不满足条件的矩形
vector<vector<Point> >contours; findContours(roiImg, contours, RETR_EXTERNAL, CHAIN_APPROX_NONE); for (int i = 0; i < contours.size(); i++) { Rect temRect = boundingRect(contours[i]); if (temRect.height < subImg.rows / 3 || temRect.height == subImg.rows) { continue; } //调整大小
Mat rectImg = Mat(roiImg, temRect); resize(rectImg, rectImg, Size(40, 50)); //与模板进行匹配
char letters = CheckImg(rectImg, 5); if (letters >= '0'&&letters <= '9' || letters == 'X') { result += letters; } } } //确定正确的 ISBN 号,来跟识别出来的 ISBN 做对比
string cmpData = ""; for (int i = 0; i < testImgFN[index].length(); i++) { if (testImgFN[index][i] >= '0'&&testImgFN[index][i] <= '9' || testImgFN[index][i] == 'X') { cmpData += testImgFN[index][i]; } } //有多余字符
if (result.length() > cmpData.length()) { string tem = result.substr(result.length() - cmpData.length()); if (tem != cmpData) { tem = result.substr(0, cmpData.length()); } result = tem; } else if (result.length() < cmpData.length()) {//有字符未被识别
int i; for (i = 0; i < result.length(); i++) { if (result[i] != cmpData[i]) { break; } } string r1 = result.substr(0, i); string r2 = result.substr(i); string r3 = ""; for (int j = 0; j < cmpData.length() - result.length();j++) { r3 += " "; } result = r1 + r3 + r2; } cout << result << endl << cmpData << endl << index << endl; //计算准确率
sunNums += cmpData.length(); for (int i = 0; i < cmpData.length(); i++) { if (result[i] == cmpData[i]) { accNums++; } } //计算正确率
if (result == cmpData) { rtNums++; cout << "Yes" << endl; } else{ cout << "No" << endl; } cout << endl; } //cout << accNums << " " << sunNums << endl;
printf("正确个数:%4.d 正确率:%f\n", rtNums, rtNums * 1.0 / testImgNums); printf("准确个数:%4.d 准确率:%f\n", accNums, accNums * 1.0 / sunNums); waitKey(0); system("pause"); }
其中 “数据集” 和 “样例” 两个文件夹均在默认路径(跟 cpp 文件放在一起)
链接:https://pan.baidu.com/s/1imCmPFalLL2Et1GeVRp7kA
提取码:m1ga