How to eliminate erroneous match(es) in matchFeatures example?

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Canberk Suat Gurel
Canberk Suat Gurel 2018 年 3 月 22 日
コメント済み: Image Analyst 2018 年 3 月 23 日
Hi all,
I have a set of images that are taken from the dash cam of a car. I am trying to find a set of corresponding points in two successive frames to estimate the direction of the car. (I will use this information to obtain the visual odometry of the car). I used the match features code given here: https://www.mathworks.com/help/vision/ref/matchfeatures.html
But sometimes I am getting wrong matches. For example:
My function:
function [matchedPoints1, matchedPoints2] = Corresponding_Points(Im1,Im2)
%%Find Corresponding Interest Points Between Pair of Images
% Find the corners.
points1 = detectHarrisFeatures(Im1);
points2 = detectHarrisFeatures(Im2);
% Extract the neighborhood features.
[features1,valid_points1] = extractFeatures(Im1,points1);
[features2,valid_points2] = extractFeatures(Im2,points2);
% Match the features.
indexPairs = matchFeatures(features1,features2);
% Retrieve the locations of the corresponding points for each image.
matchedPoints1 = valid_points1(indexPairs(:,1),:);
matchedPoints2 = valid_points2(indexPairs(:,2),:);
end
My code:
[matchedPoints1, matchedPoints2] = Corresponding_Points(image_1_gray,image_2_gray);
x1 = matchedPoints1.Location';
x2 = matchedPoints2.Location';
Note: the returned matchedPoints1 and matchedPoints2 belong to class cornerPoints . Using matchedPoints1.Location returns a matrix type single. (I am using Matlab 2016a.)
Thanks!

採用された回答

Image Analyst
Image Analyst 2018 年 3 月 22 日
Do you know the (x,y) coordinates of each? You could just throw out any pairs where the distance between them is more than, say 10% of the image width.
  2 件のコメント
Canberk Suat Gurel
Canberk Suat Gurel 2018 年 3 月 22 日
編集済み: Canberk Suat Gurel 2018 年 3 月 22 日
Thank you for your answer. Yes, I know their (x,y) coordinates. I can do something like this:
for i = 1:length(x1)
d = sqrt((x1(1,i)-x2(1,i))^2+(x1(2,i)-x2(2,i))^2)
if(d>100)
x1(:,i) = [];
x2(:,i) = [];
end
end
Can you think of a more efficient method? Getting rid of the for loop would be great.
Image Analyst
Image Analyst 2018 年 3 月 23 日
Not sure if shrinking the array at every iteration is efficient. Maybe try just keeping track of columns to keep and vectorize it and extract:
distances = sqrt((x1(1,:)-x2(1,:)).^2+(x1(2,:)-x2(2,:)).^2)
columnsToKeep = distances < 100;
x1 = x1(:, columnsToKeep);
x2 = x2(:, columnsToKeep);

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