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Distance between all points in array and delete the second point if it is less that certian value (Victorized)

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Majeed
Majeed on 12 Nov 2019
Commented: Rik on 13 Nov 2019
I have this loop to calculate the distance between all of the points in R_all array and delete the second point if the distace less that 0.002, but if I have a huge number of points like 100000 it takes long time, I need to vectorize my code, if you can help me ,, thank you in advance,,
Rx=rand(n,1)*0.2;
Ry=rand(n,1)*0.2;
Rz=rand(n,1)*0.2;
R_all=[Rx Ry Rz];
n= 1000;
Df=0.002;
while j<n
i=j+1;
while i<=n
k = norm(R_all(j,:)-R_all(i,:)); %function of distance between points
if k < 1.5*Df % check the distance between all points; should not be < 1.5 Df
R_all(i,:)=[];
n=n-1;
end
i=i+1;
end
j=j+1;
end

Accepted Answer

Rik
Rik on 13 Nov 2019
After a few runs I think this would be equivalent to your loop version. Note that it generates a matrix of size [n n 3] in one of the steps, so memory might become an issue. The pdist function would probably help, but I don't have the statistics and machine learning toolbox.
n= 1000;
Rx=rand(n,1)*0.2;
Ry=rand(n,1)*0.2;
Rz=rand(n,1)*0.2;
R_all=[Rx,Ry,Rz];
Df=0.002;
%calculate distance matrix for every point pair
A=permute(R_all,[1 3 2]);
B=permute(R_all,[3 1 2]);
dist=sqrt( sum( (A-B).^2 , 3) );
%mask the distance to the point itself and to all previous points
dist(1:(1+size(dist,1)):end)=inf;
dist(logical(triu(ones(size(dist)))))=inf;
L=dist<1.5*Df;
L=any(L,1);
R_all(L,:)=[];
  2 Comments
Rik
Rik on 13 Nov 2019
That means you're using a release prior to R2016b. If you're using an older version it is always a good idea to mention which release you are using.
In this case you can do the implicit expansion like this:
dist=sqrt( sum( bsxfun(@minus,A,B).^2 , 3) );

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More Answers (1)

Jeremy
Jeremy on 12 Nov 2019
Edited: Jeremy on 12 Nov 2019
Something like this?
n= 1000;
Rx=rand(n,1)*0.2;
Ry=rand(n,1)*0.2;
Rz=rand(n,1)*0.2;
R_all=[Rx Ry Rz];
Df=0.002;
Dr = diff(R_all,[],1); % compute differences between each row
% changing the third input from 1 to 2 makes diff compute column differences instead
[i,j,k] = find(Dr<1.5*Df); % get indices of out-of-tolerance lines
i = i + 1; % Delete the second of the lines that makes an OOT result
R_all(i,:) = []; % remove lines that are out-of-tolerance
  2 Comments
Majeed
Majeed on 13 Nov 2019
Thank you Jeremy;
this code does not find a distance as a value (magnitude) to compare with 1.5 Df. also the loop that I showed in my question was calculating each point with all other points that mean calculate the distance between point 1 with all points from 2 to 1000 then point 2 with all points from 3 to 1000 and so on. many thanks for your help.

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