android - Android中的OpenCV图像比较和相似性

Question

我是 OpenCV 学习者。我正在尝试图像比较。我用过OpenCV 2.4.13.3我有这两张图片 1.jpg 和 cam1.jpg。

当我在openCV中使用以下命令时

File sdCard = Environment.getExternalStorageDirectory();
String path1, path2;
path1 = sdCard.getAbsolutePath() + "/1.jpg";
path2 = sdCard.getAbsolutePath() + "/cam1.jpg";

FeatureDetector detector = FeatureDetector.create(FeatureDetector.ORB);
DescriptorExtractor extractor = DescriptorExtractor.create(DescriptorExtractor.BRIEF);
DescriptorMatcher matcher = DescriptorMatcher.create(DescriptorMatcher.BRUTEFORCE_HAMMING);

Mat img1 = Highgui.imread(path1);
Mat img2 = Highgui.imread(path2);

Mat descriptors1 = new Mat();
MatOfKeyPoint keypoints1 = new MatOfKeyPoint();
detector.detect(img1, keypoints1);
extractor.compute(img1, keypoints1, descriptors1);

//second image
// Mat img2 = Imgcodecs.imread(path2);
Mat descriptors2 = new Mat();
MatOfKeyPoint keypoints2 = new MatOfKeyPoint();
detector.detect(img2, keypoints2);
extractor.compute(img2, keypoints2, descriptors2);


//matcher image descriptors
MatOfDMatch matches = new MatOfDMatch();
matcher.match(descriptors1,descriptors2,matches);

// Filter matches by distance
MatOfDMatch filtered = filterMatchesByDistance(matches);

int total = (int) matches.size().height;
int Match= (int) filtered.size().height;
Log.d("LOG", "total:" + total + " Match:"+Match);

方法 filterMatchesByDistance

    static MatOfDMatch filterMatchesByDistance(MatOfDMatch matches){
    List<DMatch> matches_original = matches.toList();
    List<DMatch> matches_filtered = new ArrayList<DMatch>();

    int DIST_LIMIT = 30;
    // Check all the matches distance and if it passes add to list of filtered matches
    Log.d("DISTFILTER", "ORG SIZE:" + matches_original.size() + "");
    for (int i = 0; i < matches_original.size(); i++) {
        DMatch d = matches_original.get(i);
        if (Math.abs(d.distance) <= DIST_LIMIT) {
            matches_filtered.add(d);
        }
    }
    Log.d("DISTFILTER", "FIL SIZE:" + matches_filtered.size() + "");

    MatOfDMatch mat = new MatOfDMatch();
    mat.fromList(matches_filtered);
    return mat;
}

日志

total:122 Match:30

从日志中我们可以看出匹配是30。
但正如我们所见，两张图片具有相同的视觉元素(in)。
如何使用 openCV 获取 match=90？
如果有人可以帮助提供代码 fragment ，那就太好了。
如果使用 opencv 是不可能的那么其他是什么我们可以寻找替代方案？

Answer 1

But as we can see both images have same visual element (in).

因此，我们应该比较的不是整个图像，而是“相同的视觉元素”。如果您不比较"template"和“相机”图像本身，而是以相同的方式处理(例如，转换为二进制黑白)"template"和“相机”，则可以进一步提高 Match 值“图片。例如，尝试在两个("template"和“相机”)图像上找到蓝色(模板 Logo 的背景)方 block 并比较该方 block (感兴趣区域)。代码可能是这样的:

Bitmap bmImageTemplate = <get your template image Bitmap>;
Bitmap bmTemplate = findLogo(bmImageTemplate);  // process template image

Bitmap bmImage = <get your camera image Bitmap>;
Bitmap bmLogo = findLogo(bmImage); // process camera image same way

compareBitmaps(bmTemplate, bmLogo);

在哪里

private Bitmap findLogo(Bitmap sourceBitmap) {
    Bitmap roiBitmap = null;
    Mat sourceMat = new Mat(sourceBitmap.getWidth(), sourceBitmap.getHeight(), CvType.CV_8UC3);
    Utils.bitmapToMat(sourceBitmap, sourceMat);
    Mat roiTmp = sourceMat.clone();

    final Mat hsvMat = new Mat();
    sourceMat.copyTo(hsvMat);

    // convert mat to HSV format for Core.inRange()
    Imgproc.cvtColor(hsvMat, hsvMat, Imgproc.COLOR_RGB2HSV);

    Scalar lowerb = new Scalar(85, 50, 40);         // lower color border for BLUE
    Scalar upperb = new Scalar(135, 255, 255);      // upper color border for BLUE
    Core.inRange(hsvMat, lowerb, upperb, roiTmp);   // select only blue pixels

    // find contours
    List<MatOfPoint> contours = new ArrayList<>();
    List<Rect> squares = new ArrayList<>();
    Imgproc.findContours(roiTmp, contours, new Mat(), Imgproc.RETR_LIST, Imgproc.CHAIN_APPROX_SIMPLE);

    // find appropriate bounding rectangles
    for (MatOfPoint contour : contours) {
        MatOfPoint2f areaPoints = new MatOfPoint2f(contour.toArray());
        RotatedRect boundingRect = Imgproc.minAreaRect(areaPoints);

        double rectangleArea = boundingRect.size.area();

        // test min ROI area in pixels
        if (rectangleArea > 400) {
            Point rotated_rect_points[] = new Point[4];
            boundingRect.points(rotated_rect_points);

            Rect rect = Imgproc.boundingRect(new MatOfPoint(rotated_rect_points));
            double aspectRatio = rect.width > rect.height ?
                    (double) rect.height / (double) rect.width : (double) rect.width / (double) rect.height;
            if (aspectRatio >= 0.9) {
                squares.add(rect);
            }
        }
    }

    Mat logoMat = extractSquareMat(roiTmp, getBiggestSquare(squares));

    roiBitmap = Bitmap.createBitmap(logoMat.cols(), logoMat.rows(), Bitmap.Config.ARGB_8888);
    Utils.matToBitmap(logoMat, roiBitmap);
    return roiBitmap;
}

方法extractSquareMat()只是从整个图像中提取感兴趣区域(标志)

public static Mat extractSquareMat(Mat sourceMat, Rect rect) {
    Mat squareMat = null;
    int padding = 50;

    if (rect != null) {
        Rect truncatedRect = new Rect((int) rect.tl().x + padding, (int) rect.tl().y + padding,
                rect.width - 2 * padding, rect.height - 2 * padding);
        squareMat = new Mat(sourceMat, truncatedRect);
    }

    return squareMat ;
}

和compareBitmaps()只是你代码的包装器:

private void compareBitmaps(Bitmap bitmap1, Bitmap bitmap2) {
    Mat mat1 = new Mat(bitmap1.getWidth(), bitmap1.getHeight(), CvType.CV_8UC3);
    Utils.bitmapToMat(bitmap1, mat1);

    Mat mat2 = new Mat(bitmap2.getWidth(), bitmap2.getHeight(), CvType.CV_8UC3);
    Utils.bitmapToMat(bitmap2, mat2);

    compareMats(mat1, mat2);
}

你的代码作为方法:

private void compareMats(Mat img1, Mat img2) {
    FeatureDetector detector = FeatureDetector.create(FeatureDetector.ORB);
    DescriptorExtractor extractor = DescriptorExtractor.create(DescriptorExtractor.BRIEF);
    DescriptorMatcher matcher = DescriptorMatcher.create(DescriptorMatcher.BRUTEFORCE_HAMMING);

    Mat descriptors1 = new Mat();
    MatOfKeyPoint keypoints1 = new MatOfKeyPoint();
    detector.detect(img1, keypoints1);
    extractor.compute(img1, keypoints1, descriptors1);

    //second image
    // Mat img2 = Imgcodecs.imread(path2);
    Mat descriptors2 = new Mat();
    MatOfKeyPoint keypoints2 = new MatOfKeyPoint();
    detector.detect(img2, keypoints2);
    extractor.compute(img2, keypoints2, descriptors2);


    //matcher image descriptors
    MatOfDMatch matches = new MatOfDMatch();
    matcher.match(descriptors1,descriptors2,matches);

    // Filter matches by distance
    MatOfDMatch filtered = filterMatchesByDistance(matches);

    int total = (int) matches.size().height;
    int Match= (int) filtered.size().height;
    Log.d("LOG", "total:" + total + " Match:" + Match);
}

static MatOfDMatch filterMatchesByDistance(MatOfDMatch matches){
    List<DMatch> matches_original = matches.toList();
    List<DMatch> matches_filtered = new ArrayList<DMatch>();

    int DIST_LIMIT = 30;
    // Check all the matches distance and if it passes add to list of filtered matches
    Log.d("DISTFILTER", "ORG SIZE:" + matches_original.size() + "");
    for (int i = 0; i < matches_original.size(); i++) {
        DMatch d = matches_original.get(i);
        if (Math.abs(d.distance) <= DIST_LIMIT) {
            matches_filtered.add(d);
        }
    }
    Log.d("DISTFILTER", "FIL SIZE:" + matches_filtered.size() + "");

    MatOfDMatch mat = new MatOfDMatch();
    mat.fromList(matches_filtered);
    return mat;
}

从您的问题结果中保存的调整大小(缩放 50%)图像的结果是:

D/DISTFILTER: ORG SIZE:237
D/DISTFILTER: FIL SIZE:230
D/LOG: total:237 Match:230

注意!这是一个快速而肮脏的示例，仅用于演示给定模板的方法。

附言getBiggestSquare() 可以这样(基于面积比较):

public static Rect getBiggestSquare(List<Rect> squares) {
    Rect biggestSquare = null;

    if (squares != null && squares.size() >= 1) {
        Rect square;
        double maxArea;
        int ixMaxArea = 0;

        square = squares.get(ixMaxArea);
        maxArea = square.area();

        for (int ix = 1; ix < squares.size(); ix++) {
            square = squares.get(ix);
            if (square.area() > maxArea) {
                maxArea = square.area();
                ixMaxArea = ix;
            }
        }

        biggestSquare = squares.get(ixMaxArea);
    }

    return biggestSquare;
}