Question

我是OpenCV学习者。我正在尝试图像比较。我使用过 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

正如我们从日志匹配中看到的那样是30 但是我们可以看到两个图像都具有相同的视觉元素（in）如何使用openCV获得match = 90？如果有人可以提供代码片段，那将会很棒如果使用opencv则不可能那么另一个是什么我们可以寻找替代方案吗？

Answer 1

但是我们可以看到两个图像都具有相同的视觉元素（in）。

因此，我们应该比较不是整个图像，而是＆＃34;相同的视觉元素＆＃34;在上面。如果您不比较＆＃34;模板＆＃34;您可以更多地提高Match值。和＃34;相机＆＃34;图像本身，但处理方式相同（例如转换为二进制黑/白）＆＃34;模板＆＃34;和＃34;相机＆＃34;图片。例如，尝试在两个（＆＃34;模板＆＃34;和＃34;相机＆＃34;）图像上找到蓝色（模板徽标的背景）正方形，并比较这些正方形（感兴趣的区域）。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);

，其中

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()只从整个图像中提取感兴趣区域（徽标）

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()只为您的代码包装：

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);
}

您的代码作为方法：

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;
}

结果调整大小（缩放为50％）从您的问题结果中保存的图像是：

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

NB！这是一个快速而肮脏的例子，仅用于演示给定模板的方法。

P.S。 getBiggestSquare()可以是这样的（基于按区域比较）：

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;
}

Android中的OpenCV图像比较和相似性

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