找出两个图像是否相似 [英] Finding if two images are similar
问题描述
我试图找出两个图像与openCV使用图像匹配相似。我正在运行以下代码：
I'm trying to find out it two images are similar with openCV using image matching. I'm running the following code:
public static void match(String firstImage, String secondImage, String outputFile) {
FeatureDetector detector = FeatureDetector.create(FeatureDetector.ORB);
DescriptorExtractor descriptor = DescriptorExtractor.create(DescriptorExtractor.ORB);
DescriptorMatcher matcher = DescriptorMatcher.create(DescriptorMatcher.BRUTEFORCE_HAMMING);
Mat firstImg = Imgcodecs.imread(firstImage, Imgcodecs.CV_LOAD_IMAGE_GRAYSCALE);
MatOfKeyPoint firstKeypoints = new MatOfKeyPoint();
Mat firstDescriptors = new Mat();
detector.detect(firstImg, firstKeypoints);
descriptor.compute(firstImg, firstKeypoints, firstDescriptors);
Mat secondImg = Imgcodecs.imread(secondImage, Imgcodecs.CV_LOAD_IMAGE_GRAYSCALE);
MatOfKeyPoint secondKeypoints = new MatOfKeyPoint();
Mat secondDescriptors = new Mat();
detector.detect(secondImg, secondKeypoints);
descriptor.compute(secondImg, secondKeypoints, secondDescriptors);
MatOfDMatch matches = new MatOfDMatch();
matcher.match(firstDescriptors, secondDescriptors, matches);
float minDis = Float.MAX_VALUE;
for (int i = 0;i < matches.rows();i++) {
if (matches.toArray()[i].distance < minDis)
minDis = matches.toArray()[i].distance;
}
LinkedList<DMatch> goodMatches = new LinkedList<>();
for (int i = 0;i < matches.rows();i++) {
if (matches.toArray()[i].distance < minDis*3)
goodMatches.add(matches.toArray()[i]);
}
List<Point> pts1 = new ArrayList<Point>();
List<Point> pts2 = new ArrayList<Point>();
for(int i = 0; i<goodMatches.size(); i++){
pts1.add(firstKeypoints.toList().get(goodMatches.get(i).queryIdx).pt);
pts2.add(secondKeypoints.toList().get(goodMatches.get(i).trainIdx).pt);
}
// convertion of data types - there is maybe a more beautiful way
Mat outputMask = new Mat();
MatOfPoint2f pts1Mat = new MatOfPoint2f();
pts1Mat.fromList(pts1);
MatOfPoint2f pts2Mat = new MatOfPoint2f();
pts2Mat.fromList(pts2);
Calib3d.findHomography(pts1Mat, pts2Mat, Calib3d.RANSAC, 15, outputMask, 2000, 0.995);
// outputMask contains zeros and ones indicating which matches are filtered
LinkedList<DMatch> betterMatches = new LinkedList<DMatch>();
for (int i = 0; i < goodMatches.size(); i++) {
if (outputMask.get(i, 0)[0] != 0.0) {
betterMatches.add(goodMatches.get(i));
}
}
Mat outputImg = new Mat();
MatOfDMatch betterMatchesMat = new MatOfDMatch();
betterMatchesMat.fromList(betterMatches);
Features2d.drawMatches(firstImg, firstKeypoints, secondImg, secondKeypoints, betterMatchesMat, outputImg);
Imgcodecs.imwrite(outputFile, outputImg);
}
当图像相似时，结果如下：
When the images are similar the result looks like:
当图像不相似时，结果如下：
When the images are not similar the result looks like:
你可以看到，在第一种情况下，匹配线是平行的，因此两个图像相似是有意义的。在第二种情况下，匹配线不平行，因此图像不相似是有意义的。有没有一种标准的方法来分析这些匹配并找到在哪种情况下图像最可能相似？
You can see that in the first case the match lines are parallel so it makes sense that the two images are similar. In the second case the match lines are not parallel so it makes sense that the images are not similar. Is there a standard way to analyse these matches and to find in which case the images are most probably similar?
推荐答案
为了检查相似图像的两个视角之间的单应矩阵的适当性，我编写了这个函数。您可以根据您认为更合适的选择来选择您喜欢的内容：
For checking the appropriateness of homography matrix between two perspectives of similar images, I wrote this function. You can choose what you prefer out of the possibilities, depending on which one you find more suitable:
private static boolean check_homography(Mat homography_mat){
/* Check 1. Compute the determinant of the homography, and see if it's too close
to zero for comfort*/
if(!homography_mat.empty())
{
double Determinant = Core.determinant(homography_mat);
if (Determinant > 0.1)
return true;
else
return false;
}
else
return false;
/* Check 2. Compute its SVD, and verify that the ratio of the first-to-last
singular value is not too high (order of 1.0E7). */
Mat singularValues = new Mat();
Core.SVDecomp(homography_mat, singularValues, new Mat(), new Mat(), Core.SVD_NO_UV);
System.out.print("\n Printing the singular values of the homography");
for (int i = 0; i < singularValues.rows(); i++){
for ( int j = 0; j < singularValues.cols(); j++){
System.out.print("\n Element at ( " + i + ", " + j + " ) is " + singularValues.get(i, j)[0]);
}
}
double conditionNumber = singularValues.get(0, 0)[0] / singularValues.get(2, 0)[0];
System.out.print("\n Condition number is : " + conditionNumber);
if(conditionNumber < Math.pow(10, 7)){
System.out.print("\n Homography matrix is non-singular");
return true;
}
else{
System.out.print("\n Homography matrix is singular (or very close)");
return false;
}
/* Check 3. Check the compare absolute values at (0,0) and (0,1) with (1,1) and (1,0)
* respectively. If the two differences are close to 0, the homography matrix is
* good. (This just takes of rotation and not translation)
* */
if(Math.abs((Math.abs(homography_mat.get(0, 0)[0]) - Math.abs(homography_mat.get(1, 1)[0]))) <= 0.1){
if(Math.abs((Math.abs(homography_mat.get(0, 1)[0]) - Math.abs(homography_mat.get(1, 0)[0]))) <= 0.1){
System.out.print("\n The homography matrix is good");
return true;
}
}
else{
System.out.print("\n The homography matrix is bad");
return false;
}
return false;
/*
* Check 4: If the determinant of the top-left 2 by 2 matrix (rotation) > 0, transformation is orientation
* preserving.
* Else if the determinant is < 0, it is orientation reversing
*
* */
Determinant of the rotation mat
double det = homography_mat.get(0, 0)[0] * homography_mat.get(1,1)[0] - homography_mat.get(0, 1)[0] * homography_mat.get(1, 0)[0];
if (det < 0)
return false;
double N1 = Math.sqrt(homography_mat.get(0, 0)[0] * homography_mat.get(0, 0)[0] + homography_mat.get(1, 0)[0] * homography_mat.get(1, 0)[0]);
if (N1 > 4 || N1 < 0.1)
return false;
double N2 = Math.sqrt(homography_mat.get(0, 1)[0] * homography_mat.get(0, 1)[0] + homography_mat.get(1, 1)[0] * homography_mat.get(1, 1)[0]);
if (N2 > 4 || N2 < 0.1)
return false;
double N3 = Math.sqrt(homography_mat.get(2, 0)[0] * homography_mat.get(2, 0)[0] + homography_mat.get(2,1)[0] * homography_mat.get(2, 1)[0]);
if (N3 < 0.002)
return false;
return true;
}
注意 - 我编码了这个对于Java，OpenCV使用ORB时。我个人（有经验）我可以看一下Homography矩阵，如果它的优点与否，或多或少说， Check 1 。希望它有所帮助!!
Note - I coded this for Java, OpenCV when using ORB. I personally (with experience I guess), can look at the Homography matrix and say more or less if its good or not and hence the Check 1. Hope it helps!!
编辑
另外，正如@Micka所提到的，这取决于您的偏好，您也可以迭代沿着两个图像逐个像素地找到相似性。我在这里发布的方法用于检查单应性是否良好，但是，应该注意
即使根据图像和方法两个图像相似，你也可能得到差的单应性（前/后处理，描述符，检测器等。）
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