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How can I convert vector<Point2f> to vector<KeyPoint>?

asked 2013-11-27 02:22:46 -0500

Nihad gravatar image

updated 2013-11-27 02:46:46 -0500

berak gravatar image

I want to extract SIFT descriptor from my choosen points. How can I convert vector<point2f> to vector<keypoint>? Here is my code segment. Image resolution is 256 by 256. Thanks in advance.

int main(...)
{
.......................
Mat image = imread("test0.png",CV_LOAD_IMAGE_GRAYSCALE);
std::vector<cv::Point2f> inputs;

inputs.push_back(cv::Point2f(0,0));
inputs.push_back(cv::Point2f(0,30));
inputs.push_back(cv::Point2f(0,90));
inputs.push_back(cv::Point2f(0,120));

vector<KeyPoint> keypoints;

// Here how can I convert vector<cv::Point2f> inputs to vector<KeyPoint> keypoints?


  //Similarly, we create a smart pointer to the SIFT extractor.
  Ptr<DescriptorExtractor> featureExtractor = DescriptorExtractor::create("SIFT");

  // Compute the 128 dimension SIFT descriptor at each keypoint.
  // Each row in "descriptors" correspond to the SIFT descriptor for each keypoint
  Mat descriptors;
  featureExtractor->compute(image, keypoints, descriptors);
.........................
}
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answered 2013-11-27 03:30:10 -0500

Guanta gravatar image

Just create for each point a new Keypoint with size 1:

std::vector<cv::Keypoint> keypoints;
for( size_t i = 0; i < inputs.size(); i++ ) {
  keypoints.push_back(cv::Keypoint(inputs[i], 1.f));
}

As berak mentioned, typically this doesn't make sense, since your descriptor is neither scale nor rotational invariant any more.

However in several circumstances it is totally valid! E.g. the combination of of a dense grid sampling + SIFT is typically used for bag of words approaches. The dense grid sampling over the image is nothing else than a Keypoint of size 1 at each step size (e.g. 5) location in the image (note that you don't need to implement it yourself, since OpenCV has the detector "Dense" which does exactly this for you).

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oh, ofc. thanks for highlighting this !

berak gravatar imageberak ( 2013-11-27 03:36:02 -0500 )edit
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answered 2013-11-27 02:46:05 -0500

berak gravatar image

updated 2013-11-27 02:47:26 -0500

"I want to extract SIFT descriptor from my choosen points." - No, you can't choose the points yourself, instead leave it to SIFT:

Ptr<FeatureDetector> detector  = FeatureDetector::create( "SIFT" );
vector<KeyPoint> keypoints;
detector->detect( img, keypoints );
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Thank u. Is this possible to convert vector<Point2f> to vector<KeyPoint>? Neglecting the issue related to SIFT extractor. I have tried in following way, but failed. vector<KeyPoint>keypoints;

KeyPoint::convert(inputs,keypoints, 1, 1, 0, -1);

Nihad gravatar imageNihad ( 2013-11-27 02:55:31 -0500 )edit

not really.

(yes, you could construct a new Keypoint for each Point, and push it into a vector, but it does not make any sense, since you won't be able to come up with sensible values for anything besides the position)

again, what you want is not feasible.

berak gravatar imageberak ( 2013-11-27 03:00:40 -0500 )edit

Have you read the paper from Lowe concerning SIFT? If you try to understand the properties of the lineare scale space, you will understand on your own why your idea is not as good as you may think!

JohannesZ gravatar imageJohannesZ ( 2013-11-27 05:51:20 -0500 )edit
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Asked: 2013-11-27 02:22:46 -0500

Seen: 13,475 times

Last updated: Nov 27 '13