image - 关键点检测与图像拼接

Question

] 2

所以如下图所示，我在图像上检测到了关键点，但是包裹透视后的输出图像忽略了左侧的第一张图像，不知道为什么!

    import numpy as np
    import imutils
    import cv2

class Stitcher:
def __init__(self):
    # determine if we are using OpenCV v3.X
    self.isv3 = imutils.is_cv3()

def stitch(self, imageA,imageB, ratio=0.75, reprojThresh=10.0,
    showMatches=False):
    # unpack the images, then detect keypoints and extract
    # local invariant descriptors from them
    #(imageB, imageA) = images
    (kpsA, featuresA) = self.detectAndDescribe(imageA)
    (kpsB, featuresB) = self.detectAndDescribe(imageB)

    # match features between the two images
    M = self.matchKeypoints(kpsA, kpsB,
        featuresA, featuresB, ratio, reprojThresh)

    # if the match is None, then there aren't enough matched
    # keypoints to create a panorama
    if M is None:
        return None

    # otherwise, apply a perspective warp to stitch the images
    # together
    (matches, H, status) = M
    #print(M)
    #print(matches)
    #print(H)
    #print(status)
    #cv2.imwrite('intermediate.jpg',matches)
    result = cv2.warpPerspective(imageA, H,
        (imageA.shape[1] + imageB.shape[1], imageA.shape[0]))
    result[0:imageB.shape[0], 0:imageB.shape[1]] = imageB
    #cv2.imshow('intermediate',result)

    # check to see if the keypoint matches should be visualized
    if showMatches:
        vis = self.drawMatches(imageA, imageB, kpsA, kpsB, matches,
            status)

        # return a tuple of the stitched image and the
        # visualization
        return (result, vis)

    # return the stitched image
    return result

def detectAndDescribe(self, image):
    # convert the image to grayscale
    gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)

    # check to see if we are using OpenCV 3.X
    if self.isv3:
        # detect and extract features from the image
        #SIFT Algorithm
        descriptor = cv2.xfeatures2d.SIFT_create()
        #SURF Algorithm
        #descriptor = cv2.xfeatures2d.SURF_create()# 400 is hesian threshold, optimum values should be around 300-500
        #upright SURF: faster and can be used for panorama stiching i.e our case.
        #descriptor.upright = True
        print(descriptor.descriptorSize())
        (kps, features) = descriptor.detectAndCompute(image, None)
        print(len(kps),features.shape)

    # otherwise, we are using OpenCV 2.4.X
    else:
        # detect keypoints in the image
        detector = cv2.FeatureDetector_create("SIFT")
        kps = detector.detect(gray)

        # extract features from the image
        extractor = cv2.DescriptorExtractor_create("SIFT")
        (kps, features) = extractor.compute(gray, kps)

    # convert the keypoints from KeyPoint objects to NumPy
    # arrays
    kps = np.float32([kp.pt for kp in kps])

    # return a tuple of keypoints and features
    #print("features",features)
    return (kps, features)

def matchKeypoints(self, kpsA, kpsB, featuresA, featuresB,
    ratio, reprojThresh):
    # compute the raw matches and initialize the list of actual
    # matches
    matcher = cv2.DescriptorMatcher_create("BruteForce")
    rawMatches = matcher.knnMatch(featuresA, featuresB, 2)
    matches = []

    # loop over the raw matches
    for m in rawMatches:
        # ensure the distance is within a certain ratio of each
        # other (i.e. Lowe's ratio test)
        if len(m) == 2 and m[0].distance < m[1].distance * ratio:
            matches.append((m[0].trainIdx, m[0].queryIdx))
    print(len(matches))

    # computing a homography requires at least 4 matches
    if len(matches) > 4:
        # construct the two sets of points
        ptsA = np.float32([kpsA[i] for (_, i) in matches])
        ptsB = np.float32([kpsB[i] for (i, _) in matches])

        # compute the homography between the two sets of points
        (H, status) = cv2.findHomography(ptsA, ptsB, cv2.RANSAC,
            reprojThresh)

        # return the matches along with the homograpy matrix
        # and status of each matched point
        return (matches, H, status)

    # otherwise, no homograpy could be computed
    return None

def drawMatches(self, imageA, imageB, kpsA, kpsB, matches, status):
    # initialize the output visualization image
    (hA, wA) = imageA.shape[:2]
    (hB, wB) = imageB.shape[:2]
    vis = np.zeros((max(hA, hB), wA + wB, 3), dtype="uint8")
    vis[0:hA, 0:wA] = imageA
    vis[0:hB, wA:] = imageB

    # loop over the matches
    for ((trainIdx, queryIdx), s) in zip(matches, status):
        # only process the match if the keypoint was successfully
        # matched
        if s == 1:
            # draw the match
            ptA = (int(kpsA[queryIdx][0]), int(kpsA[queryIdx][1]))
            ptB = (int(kpsB[trainIdx][0]) + wA, int(kpsB[trainIdx][1]))
            cv2.line(vis, ptA, ptB, (0, 255, 0), 1)

    # return the visualization
    return vis

以上是用于关键点检测和拼接的代码，

另一个问题是否有人可以帮助我进行垂直图像拼接而不是旋转图像和执行水平拼接。

非常感谢 !



我更改了代码并使用@Alexander 的 padtransf.warpPerspectivePadded 函数来执行包装和混合!你能帮我获得输出图像的照明均匀吗？

Answer 1

我自己也有这个问题。如果我没记错的话，您使用的是 this博客作为引用。
问题是 warpPerspective关于这条线:

result = cv2.warpPerspective(imageA, H,
        (imageA.shape[1] + imageB.shape[1], imageA.shape[0]))
    result[0:imageB.shape[0], 0:imageB.shape[1]] = imageB

这种方法是全方位的。我的意思是，您只是通过根据 .shape[0] 表示的宽度和高度替换像素值来将 imageA 拼接到 imageB 上。和 .shape[1] .我在 C++ 中解决了这个问题，因此没有要显示的 python 代码，但可以让你了解必须做的事情。

获取您正在使用的每个图像的四个角。

获取在步骤 1 中找到的每个图像的最小和最大角点。

创建一个垫子“HTR”，用于映射图像一，使其与已经变形的图像二一致。 HTR.at(0,2) 代表 mats 3x3 矩阵中的一个位置。 Numpy 可能是您需要在这里使用的。

Mat Htr = Mat::eye(3,3,CV_64F);
    if (min_x < 0){
        max_x = image2.size().width - min_x;
        Htr.at<double>(0,2)= -min_x;
    }
    if (min_y < 0){
        max_y = image2.size().height - min_y;
        Htr.at<double>(1,2)= -min_y;
    }

对每个图像的四个角进行透视变换，以查看它们在空间中的最终位置。

perspectiveTransform(vector<Point2f> fourPointImage1, vector<Point2f> image1dst, Htr*homography);
perspectiveTransform(vector<Point2f> fourPointImage2, vector<Point2f> image2dst, Htr);

从 image1dst 获取最小值和最大值四个角和iamge2dst四个角。

获取 image1dst 的最小值和最大值和 iamge2dst并使用创建一个新的blank image正确大小以保存最终拼接的图像。

这次重复步骤 3 过程以确定 translation需要调整每个图像的四个角以确保将它们移动到 blank image 的虚拟空间中

最后将您找到/制作的所有单应性放入实际图像中。

warpPerspective(image1, blankImage, (translation*homography),result.size(), INTER_LINEAR,BORDER_CONSTANT,(0));
warpPerspective(image2, image2Updated, translation, result.size(), INTER_LINEAR, BORDER_CONSTANT,   (0));

最终目标和结果是确定图像将被扭曲到哪里，这样您就可以制作一个空白图像来保存整个拼接图像，这样就不会裁剪任何内容。只有完成所有预处理后，您才能真正将图像拼接在一起。我希望这会有所帮助，如果您有问题，请大声疾呼。