我在android平台上使用openCV库。 我已经成功检测到图像中最大的矩形,但由于我的应用程序将用于扫描目的,我也希望获得透视更改功能。
我知道,如何应用perspectiveTransform和warpPerspectiveTransform,但为此我需要使用矩形的角来作为源点。
考虑到我们拥有与Rect对象关联的第一个角(左上角)和宽度/高度的坐标这一事实似乎很容易找到角落,但问题是,对于旋转的矩形(通常的boundingRect但是侧面)不平行于轴),这些值是非常不同的。在这种情况下,它存储对应于另一个矩形的值,该矩形的边平行于轴并覆盖旋转的矩形,这使我无法检测实际矩形的角。
此外,我想对这两种算法进行比较,以便从图像中检测纸张。
Canny edge - >最大轮廓 - >最大的矩形 - >找到角落 - >透视变化
Canny edge->霍夫线 - >线的交叉点 - >透视变化
如果我们有一个Rect对象,如何获得该矩形的所有角落,我想问的是什么?
提前致谢。
答案 0 :(得分:11)
我很高兴回答我的问题!这很容易,但是当你刚开始使用不那么合适的相关文档的时候就会发生这种情况。
我正在努力寻找一个普通矩形的角落,这个角落在openCV的实现中没有定义,因此几乎是不可能的。
我遵循stackoverflow上的标准代码进行最大的Square检测。使用approxCurve本身可以轻松找到角落。
//将图像转换为黑白
Imgproc.cvtColor(imgSource, imgSource, Imgproc.COLOR_BGR2GRAY);
//convert the image to black and white does (8 bit)
Imgproc.Canny(imgSource, imgSource, 50, 50);
//apply gaussian blur to smoothen lines of dots
Imgproc.GaussianBlur(imgSource, imgSource, new org.opencv.core.Size(5, 5), 5);
//find the contours
List<MatOfPoint> contours = new ArrayList<MatOfPoint>();
Imgproc.findContours(imgSource, contours, new Mat(), Imgproc.RETR_LIST, Imgproc.CHAIN_APPROX_SIMPLE);
double maxArea = -1;
int maxAreaIdx = -1;
Log.d("size",Integer.toString(contours.size()));
MatOfPoint temp_contour = contours.get(0); //the largest is at the index 0 for starting point
MatOfPoint2f approxCurve = new MatOfPoint2f();
MatOfPoint largest_contour = contours.get(0);
//largest_contour.ge
List<MatOfPoint> largest_contours = new ArrayList<MatOfPoint>();
//Imgproc.drawContours(imgSource,contours, -1, new Scalar(0, 255, 0), 1);
for (int idx = 0; idx < contours.size(); idx++) {
temp_contour = contours.get(idx);
double contourarea = Imgproc.contourArea(temp_contour);
//compare this contour to the previous largest contour found
if (contourarea > maxArea) {
//check if this contour is a square
MatOfPoint2f new_mat = new MatOfPoint2f( temp_contour.toArray() );
int contourSize = (int)temp_contour.total();
MatOfPoint2f approxCurve_temp = new MatOfPoint2f();
Imgproc.approxPolyDP(new_mat, approxCurve_temp, contourSize*0.05, true);
if (approxCurve_temp.total() == 4) {
maxArea = contourarea;
maxAreaIdx = idx;
approxCurve=approxCurve_temp;
largest_contour = temp_contour;
}
}
}
Imgproc.cvtColor(imgSource, imgSource, Imgproc.COLOR_BayerBG2RGB);
sourceImage =Highgui.imread(Environment.getExternalStorageDirectory().
getAbsolutePath() +"/scan/p/1.jpg");
double[] temp_double;
temp_double = approxCurve.get(0,0);
Point p1 = new Point(temp_double[0], temp_double[1]);
//Core.circle(imgSource,p1,55,new Scalar(0,0,255));
//Imgproc.warpAffine(sourceImage, dummy, rotImage,sourceImage.size());
temp_double = approxCurve.get(1,0);
Point p2 = new Point(temp_double[0], temp_double[1]);
// Core.circle(imgSource,p2,150,new Scalar(255,255,255));
temp_double = approxCurve.get(2,0);
Point p3 = new Point(temp_double[0], temp_double[1]);
//Core.circle(imgSource,p3,200,new Scalar(255,0,0));
temp_double = approxCurve.get(3,0);
Point p4 = new Point(temp_double[0], temp_double[1]);
// Core.circle(imgSource,p4,100,new Scalar(0,0,255));
List<Point> source = new ArrayList<Point>();
source.add(p1);
source.add(p2);
source.add(p3);
source.add(p4);
Mat startM = Converters.vector_Point2f_to_Mat(source);
Mat result=warp(sourceImage,startM);
return result;
以及用于透视变换的函数如下:
public Mat warp(Mat inputMat,Mat startM) {
int resultWidth = 1000;
int resultHeight = 1000;
Mat outputMat = new Mat(resultWidth, resultHeight, CvType.CV_8UC4);
Point ocvPOut1 = new Point(0, 0);
Point ocvPOut2 = new Point(0, resultHeight);
Point ocvPOut3 = new Point(resultWidth, resultHeight);
Point ocvPOut4 = new Point(resultWidth, 0);
List<Point> dest = new ArrayList<Point>();
dest.add(ocvPOut1);
dest.add(ocvPOut2);
dest.add(ocvPOut3);
dest.add(ocvPOut4);
Mat endM = Converters.vector_Point2f_to_Mat(dest);
Mat perspectiveTransform = Imgproc.getPerspectiveTransform(startM, endM);
Imgproc.warpPerspective(inputMat,
outputMat,
perspectiveTransform,
new Size(resultWidth, resultHeight),
Imgproc.INTER_CUBIC);
return outputMat;
}
答案 1 :(得分:2)
1. find the coordinate of left ,right, top,bottom point of the rect
2. min_x = min(left.x right.x,top.x,bottom.x)
min_y = min(left.y right.y,top.y,bottom.y)
max_x = max(left.x right.x,top.x,bottom.x)
max_y = max(left.y right.y,top.y,bottom.y)
3. the corners's coordinate is the point with the min_x, min_y ,max_x, max_y