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我想从我指定的关键点计算 SURF 特征。我正在使用 OpenCV 的 Python 包装器。以下是我尝试使用的代码，但我在任何地方都找不到工作示例。

surf = cv2.SURF()
keypoints, descriptors = surf.detect(np.asarray(image[:,:]),None,useProvidedKeypoints = True)

如何指定此函数要使用的关键点？

类似的，未回答的问题： cvExtractSURF don't work when useProvidedKeypoints = true

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3 回答 3

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如果我正确理解 Python 绑定的源代码，则 C++ 接口中存在的“关键点”参数永远不会在 Python 绑定中使用。所以我冒着风险，用当前的绑定做你想做的事情是不可能的。一个可能的解决方案是编写自己的绑定。我知道这不是你希望的答案...

于 2012-08-02T08:59:53.897 回答
2

尝试为此使用 cv2.DescriptorMatcher_create 。

例如，在下面的代码中，我使用的是 pylab，但你可以得到消息；）

它使用 GFTT 计算关键点，然后使用 SURF 描述符和蛮力匹配。每个代码部分的输出显示为标题。

%pylab inline
import cv2
import numpy as np

img = cv2.imread('./img/nail.jpg')
gray= cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
imshow(gray,  cmap=cm.gray)

输出类似于http://i.stack.imgur.com/8eOTe.png

（对于这个例子，我将作弊并使用相同的图像来获取关键点和描述符）。

img1 = gray
img2 = gray
detector = cv2.FeatureDetector_create("GFTT")
descriptor = cv2.DescriptorExtractor_create("SURF")
matcher = pt1=(int(k1[m.queryIdx].pt[0]),int(k1[m.queryIdx].pt[1]))("FlannBased")

# detect keypoints
kp1 = detector.detect(img1)
kp2 = detector.detect(img2)

print '#keypoints in image1: %d, image2: %d' % (len(kp1), len(kp2))

image1中的关键点：1000，image2：1000

# descriptors
k1, d1 = descriptor.compute(img1, kp1)
k2, d2 = descriptor.compute(img2, kp2)

print '#Descriptors size in image1: %s, image2: %s' % ((d1.shape), (d2.shape))

image1 中的描述符大小：(1000, 64)，image2: (1000, 64)

# match the keypoints
matches = matcher.match(d1,d2)

# visualize the matches
print '#matches:', len(matches)
dist = [m.distance for m in matches]

print 'distance: min: %.3f' % min(dist)
print 'distance: mean: %.3f' % (sum(dist) / len(dist))
print 'distance: max: %.3f' % max(dist)

匹配数：1000

距离：分钟：0.000

距离：平均值：0.000

距离：最大：0.000

# threshold: half the mean
thres_dist = (sum(dist) / len(dist)) * 0.5 + 0.5

# keep only the reasonable matches
sel_matches = [m for m in matches if m.distance < thres_dist]

print '#selected matches:', len(sel_matches)

选择匹配：1000

#Plot
h1, w1 = img1.shape[:2]
h2, w2 = img2.shape[:2]
view = zeros((max(h1, h2), w1 + w2, 3), uint8)
view[:h1, :w1, 0] = img1
view[:h2, w1:, 0] = img2
view[:, :, 1] = view[:, :, 0]
view[:, :, 2] = view[:, :, 0]

for m in sel_matches:
    # draw the keypoints
    # print m.queryIdx, m.trainIdx, m.distance
    color = tuple([random.randint(0, 255) for _ in xrange(3)])
    pt1=(int(k1[m.queryIdx].pt[0]),int(k1[m.queryIdx].pt[1]))
    pt2=(int(k2[m.queryIdx].pt[0]+w1),int(k2[m.queryIdx].pt[1]))
    cv2.line(view,pt1,pt2,color)

输出是这样的http://i.stack.imgur.com/8CqrJ.png

于 2013-10-30T11:03:33.130 回答
0

前面提到的如何做到这一点的例子Mahotas：

import mahotas
from mahotas.features import surf
import numpy as np


def process_image(imagename):
    '''Process an image and returns descriptors and keypoints location'''
    # Load the images
    f = mahotas.imread(imagename, as_grey=True)
    f = f.astype(np.uint8)

    spoints = surf.dense(f, spacing=12, include_interest_point=True)
    # spoints includes both the detection information (such as the position
    # and the scale) as well as the descriptor (i.e., what the area around
    # the point looks like). We only want to use the descriptor for
    # clustering. The descriptor starts at position 5:
    desc = spoints[:, 5:]
    kp = spoints[:, :2]

    return kp, desc
于 2013-06-19T21:09:51.250 回答