根据矩阵的二维相关系数进行OCR识别

Posted 2022-12-11 lc__________

我想通过简单的模板匹配来进行图像识别。

把预处理好的字符图片，分别与A到Z的样本图片进行模板匹配。结果最大的表明相关性最大，就可以识别字符图片了。

在实际应用中，我用了openCV的matchTemplate()函数，但是未达到我想要点的效果。

matchTemplate()的功能是在图像中搜索出指定的模板，如果模板是从待搜索的图像中截取出来的，会有很好的效果。但是如果模板不是待搜素图像的一部分，似乎达不到我想要的效果。

在尝试了matlab的corr2()后，发现corr2能很好的解决我的问题。

double Card::corr2(Mat matA, Mat matB)
	//计算两个相同大小矩阵的二维相关系数
	double corr2 = 0;

	double Amean2 = 0; 
	double Bmean2 = 0;
	for (int m = 0; m < matA.rows; m++)   
		uchar* dataA = matA.ptr<uchar>(m);  
		uchar* dataB = matB.ptr<uchar>(m); 
		for (int n = 0; n < matA.cols;n++)   			
			Amean2 = Amean2 + dataA[n];  
			Bmean2 = Bmean2 + dataB[n];
		                    
	
	Amean2 = Amean2 / (matA.rows * matA.cols);
	Bmean2 = Bmean2 / (matB.rows * matB.cols);

	double Cov = 0;
	double Astd = 0;
	double Bstd = 0;
	for (int m = 0; m < matA.rows; m++)   
		uchar* dataA = matA.ptr<uchar>(m);  
		uchar* dataB = matB.ptr<uchar>(m); 
		for (int n = 0; n < matA.cols;n++)  
			//协方差
			Cov = Cov + (dataA[n] - Amean2) * (dataB[n] - Bmean2);
			//A的方差
			Astd = Astd + (dataA[n] - Amean2) * (dataA[n] - Amean2);
			//B的方差
			Bstd = Bstd + (dataB[n] - Bmean2) * (dataB[n] - Bmean2);
		                    
	
	corr2 = Cov / (sqrt(Astd * Bstd));

	return corr2;

//待搜索图像
Mat srcImage = imread("M:/图像处理实验/验证码/byx001.bmp",1);
Mat resizeMat = Mat::zeros(25, 25, CV_8UC3);
//缩放为25*25的矩阵。因为要相关匹配的模板大小为25*25
resize(srcImage,resizeMat,resizeMat.size());

//相关匹配	
double ccorrVal[26] = 0;

double max = 0;
int count = 0;
for (int m = 0; m < 26; m++)
	char recogPath[100] = 1;
	strcpy(recogPath,"M://图像处理实验//验证码//大写字母//");
	char num[2] = 1;
	num[0] = 65 + m;
	strcat(recogPath, num);
	strcat(recogPath,".bmp");

	Mat img_display;
	resizeMat.copyTo( img_display );
	Mat std = imread(recogPath,0);

	Mat resizeMatSTD = Mat::zeros(25, 25, CV_8UC3);
	resize(std,resizeMatSTD,resizeMatSTD.size());

	adaptiveThreshold(resizeMatSTD ,resizeMatSTD ,255 ,ADAPTIVE_THRESH_MEAN_C ,THRESH_BINARY,5,1);
				
	corr2Val[m] = corr2(resizeMatSTD,img_display);
				
	if (max <= corr2Val[m])
		max = corr2Val[m];
		count = m;
		

char pathname[100]=1;
strcpy(pathname,"M://图像处理实验//验证码//test//字符_");
char num[10];
_itoa(i, num, 10);
strcat(pathname, num);


char C[2] = 1;
C[0] = 65 + count;
strcat(pathname, C);
strcat(pathname,".bmp");
imwrite(pathname, resizeMat);

字符模板：

识别结果输出：