Find matching points from two coordinate systems
27 次查看(过去 30 天)
显示 更早的评论
Hello,
I have 10 points with x and y coordinate (cell locations imaged with microscope). I have used two different methods to image these cells, so I have two sets of coordinates that could be rotated, sheared, shifted, mirrored to each other. I would like to assign each point from one coordinate system to the matching point in the other coordinate system. How can I do that? In theory, some kind of transformation optimizing minimum distance between the points would be desirable.
In the end I would like to assign properties I obtained from the cells in one method to properties obtained with the other method.
Thank you.
7 个评论
采纳的回答
Matt J
2019-10-8
编辑:Matt J
2019-10-8
If the shear component of the deformation isn't too strong, the matchpoints function defined below might work. It will sort the rows of B to correspond with A and also return the corresponding permutation indices. It relies on absor, mentioned by Bruno, which you will have to download
function [permIndices,Bsorted]=matchpoints(A,B)
%
%IN:
%
% A: an Nx2 matrix of points
% B: an Nx2 matrix of points from A transformed and unordered
%
%OUT:
%
% permIndices: permutation indices of rows of B thought to match A
% Bsorted: the Nx2 permuted version of B
La=landmarks(A);
Lb=landmarks(B);
B3=B.'; B3(3,:)=0;
reg=absor( Lb,La,'doScale',1);
C3=(reg.s*reg.R)*B3+reg.t;
C=C3(1:2,:).';
[~,permIndices]=pdist2(C,A,'euclidean','Smallest',1);
if nargout>1
Bsorted=B(permIndices,:);
end
function L=landmarks(P)
G=pdist2(P,P); G(~G)=nan;
[i,j]=find( G==min(G(:)) ,1);
I=P(i,:);
J=P(j,:);
K=mean(P,1);
if norm(I-K)<norm(J-K)
[I,J]=deal(J,I);
end
L=[I;J;K].';
L(3,:)=0;
end
end
Applying it to your example data, I obtain,
A = [376 455;421 489;465 537;353 512;355 535;329 571;377 593;417 598;482 575;355 634];
B = [168 88;107 138;69 194;126 229;163 232;199 267;272 239;228 210;235 155;215 68];
[permIndices,Bsorted]=matchpoints(A,B)
plot(graph(1:10,1:10),'EdgeColor','none','XData',A(:,1),'YData',A(:,2));
hold on
plot(graph(1:10,1:10),'EdgeColor','none','XData',Bsorted(:,1),'YData',Bsorted(:,2));
hold off
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/241745/image.png)
8 个评论
Matt J
2020-5-2
编辑:Matt J
2020-5-9
Here is a modification of the code that works on your example. You have a very large outlier in your coordinates, however, so it can't always be guaranteed that such cases would be robustly handled. Note that the code uses the FEX file minL1intlin.m (Download).
function [permIndices,Bsorted]=matchpoints(A,B)
%
%IN:
%
% A: an Nx2 matrix of points
% B: an Nx2 matrix of points from A transformed and unordered
%
%OUT:
%
% permIndices: permutation indices of rows of B thought to match A
% Bsorted: the Nx2 permuted version of B
La=landmarks(A);
Lb=landmarks(B);
B3=B.'; B3(3,:)=0;
reg=absor( Lb,La,'doScale',1);
C3=(reg.s*reg.R)*B3+reg.t;
C=C3(1:2,:).';
N=size(A,1);
e=ones(1,N);
E=speye(N);
Aeq=[ kron(E,e) ; kron(e,E) ]; beq=[e,e].';
Q=kron(E,C.');
d=reshape(A.',[],1);
lb=zeros(1,N^2);
ub=ones(1,N^2);
intcon=1:N^2;
P=minL1intlin(Q,d,intcon,[],[],Aeq,beq,lb,ub);
P=round(reshape(P,N,[]));
permIndices=(1:N)*P;
if nargout>1
Bsorted=B(permIndices,:);
end
function L=landmarks(P)
G=pdist2(P,P); G(~G)=nan;
[i,j]=find( G==min(G(:)) ,1);
I=P(i,:);
J=P(j,:);
K=mean(P,1);
if norm(I-K)<norm(J-K)
[I,J]=deal(J,I);
end
L=[I;J;K].';
L(3,:)=0;
end
end
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/289620/image.png)
更多回答(1 个)
Bruno Luong
2019-10-8
Look for the literature of image registration.
For simple rotation/scaling/translation you can use Matt J's submission
For more complex deformation, you need to apply spline deformation
There are a bunch of intermediate method for camera which take into account for camera cushion distortion or higher order. Pick one that is suitable for your need.
2 个评论
另请参阅
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!