Ale*_*lex 6 python opencv image-processing sift feature-detection
我试图用我自己的图像来遵循本教程。但是,我得到的结果并不完全符合我的预期。我在这里遗漏了什么,还是 SIFT 在这种情况下不是一个足够好的解决方案?多谢。
import numpy as np
import cv2
from matplotlib import pyplot as plt
MIN_MATCH_COUNT = 10
img1 = cv2.imread('Q/IMG_1192.JPG', 0) # queryImage
img2 = cv2.imread('DB/IMG_1208-1000.jpg', 0) # trainImage
# Initiate SIFT detector
sift = cv2.xfeatures2d.SIFT_create()
# find the keypoints and descriptors with SIFT
kp1, des1 = sift.detectAndCompute(img1,None)
kp2, des2 = sift.detectAndCompute(img2,None)
FLANN_INDEX_KDTREE = 0
index_params = dict(algorithm = FLANN_INDEX_KDTREE, trees = 5)
search_params = dict(checks = 50)
flann = cv2.FlannBasedMatcher(index_params, search_params)
matches = flann.knnMatch(des1, des2, k=2)
# store all the good matches as per Lowe's ratio test.
good = []
for m,n in matches:
if m.distance < 0.7*n.distance:
good.append(m)
if len(good)>MIN_MATCH_COUNT:
src_pts = np.float32([ kp1[m.queryIdx].pt for m in good ]).reshape(-1,1,2)
dst_pts = np.float32([ kp2[m.trainIdx].pt for m in good ]).reshape(-1,1,2)
M, mask = cv2.findHomography(src_pts, dst_pts, cv2.RANSAC,5.0)
matchesMask = mask.ravel().tolist()
h,w, = img1.shape
pts = np.float32([ [0,0],[0,h-1],[w-1,h-1],[w-1,0] ]).reshape(-1,1,2)
dst = cv2.perspectiveTransform(pts,M)
img2 = cv2.polylines(img2,[np.int32(dst)],True,255,3, cv2.LINE_AA)
else:
print ("Not enough matches are found - %d/%d" % (len(good),MIN_MATCH_COUNT))
matchesMask = None
draw_params = dict(matchColor = (0,255,0), # draw matches in green color
singlePointColor = None,
matchesMask = matchesMask, # draw only inliers
flags = 2)
img3 = cv2.drawMatches(img1,kp1,img2,kp2,good,None,**draw_params)
plt.imshow(img3, 'gray'),plt.show()
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为了评估问题是否来自 SIFT 描述符,我建议您使用其他描述符,例如cv2.xfeatures2d.VGG_create()或cv2.BRISK_create()。另请注意,cv2.xfeatures2d.matchGMS即使使用 SIFT 描述符,它也可能会给出更好的结果。
根据我个人的经验,在 SIFT 算法的应用中证明缺乏准确性的可能原因之一是对梯度反转的敏感性。SIFT 描述符确实包含关键点周围归一化梯度方向的量化直方图。问题是,如果区域中的强度从较亮的像素移动到较暗的像素(例如 255->127),则梯度方向将不同于强度从较暗的像素移动到较亮的像素(例如 127->255)。