opencv 识别长方形_利用opencv识别并提取图片中的矩形
这次是利用opencv来识别图片中的矩形。
其中遇到的问题主要是识别轮廓时矩形内部的形状导致轮廓不闭合。
过程如下:
**1. 对输入灰度图片进行高斯滤波
2. 做灰度直方图,提取阈值,做二值化处理
3. 提取图片轮廓
4. 识别图片中的矩形
5. 提取图片中的矩形**
1.对输入灰度图片进行高斯滤波
cv::Mat src = cv::imread("F:\\t13.bmp",CV_BGR2GRAY);
cv::Mat hsv;
GaussianBlur(src,hsv,cv::Size(5,5),0,0);
2.做灰度直方图,提取阈值,做二值化处理
由于给定图片,背景是黑色,矩形背景色为灰色,矩形中有些其他形状为白色,可以参考为:
提取轮廓时,矩形外部轮廓并未闭合。因此,我们需要对整幅图做灰度直方图,找到阈值,进行二值化处理。即令像素值(黑色)小于阈值的,设置为0(纯黑色);令像素值(灰色和白色)大于阈值的,设置为255(白色)
// Quantize the gray scale to 30 levels
int gbins = 16;
int histSize[] = {gbins};
// gray scale varies from 0 to 256
float granges[] = {0,256};
const float* ranges[] = { granges };
cv::MatND hist;
// we compute the histogram from the 0-th and 1-st channels
int channels[] = {0};
//calculate hist
calcHist( &hsv, 1, channels, cv::Mat(), // do not use mask
hist, 1, histSize, ranges,
true, // the histogram is uniform
false );
//find the max value of hist
double maxVal=0;
minMaxLoc(hist, 0, &maxVal, 0, 0);
int scale = 20;
cv::Mat histImg;
histImg.create(500,gbins*scale,CV_8UC3);
//show gray scale of hist image
for(int g=0;g
float binVal = hist.at(g,0);
int intensity = cvRound(binVal*255);
rectangle( histImg, cv::Point(g*scale,0),
cv::Point((g+1)*scale - 1,binVal/maxVal*400),
CV_RGB(0,0,0),
CV_FILLED );
}
cv::imshow("histImg",histImg);
//threshold processing
cv::Mat hsvRe;
threshold( hsv, hsvRe, 64, 255,cv::THRESH_BINARY);
3.提取图片轮廓
为了识别图片中的矩形,在识别之前还需要提取图片的轮廓。在经过滤波、二值化处理后,轮廓提取后的效果比未提取前的效果要好很多。
vector approx;
for (size_t i = 0; i < contours.size(); i++)
{
approxPolyDP(Mat(contours[i]), approx,
arcLength(Mat(contours[i]), true)*0.02, true);
if (approx.size() == 4 &&
fabs(contourArea(Mat(approx))) > 1000 &&
isContourConvex(Mat(approx)))
{
double maxCosine = 0;
for( int j = 2; j < 5; j++ )
{
double cosine = fabs(angle(approx[j%4], approx[j-2], approx[j-1]));
maxCosine = MAX(maxCosine, cosine);
}
if( maxCosine < 0.3 )
squares.push_back(approx);
}
}
5.提取图片中的矩形
由于图片中矩形倾斜角度不太大,所以没有做倾斜校正这步操作
这是主函数提取部分代码
//get rect from image
std::vector compression_params;
compression_params.push_back(CV_IMWRITE_PNG_COMPRESSION);
compression_params.push_back(9);
for(int i=0;i
int rect[4],*tmp;
tmp = findRectInfo(squares[i]);
for(int j=0;j<4;j++)
rect[j] = *(tmp+j);
cv::Rect roi(rect[1],rect[0],rect[3],rect[2]);
cv::Mat roi_of_image = src(roi);
char * filename = new char[100];
sprintf(filename,"F:\\vs\\ConsoleApplication2\\resultPic_t35\\%i.png",i);
cv::imwrite(filename,roi_of_image,compression_params);
}
这是计算每个矩形的左上角点x,y坐标和长宽值
int* findRectInfo(std::vector<:point> rect)
{
int rectInfo[4] = {0};
int x[4]= {0},y[4]= {0};
int maxX = 0,maxY = 0,minX = 2000,minY = 2000;
//get the rect points
for(int i=0;i<4;i++){
x[i] = rect[i].x;
y[i] = rect[i].y;
if(maxX
maxX = x[i];
if(maxY
maxY = y[i];
if(minX>x[i])
minX = x[i];
if(minY>y[i])
minY = y[i];
}
rectInfo[0] = minY;
rectInfo[1] = minX;
rectInfo[2] = maxY - minY;
rectInfo[3] = maxX - minX;
return rectInfo;
}
识别并提取结果为:

更多推荐


所有评论(0)