Distance between all points in array and delete the second point if it is less that certian value (Victorized)
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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
0 件のコメント
採用された回答
Rik
2019 年 11 月 13 日
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 件のコメント
Rik
2019 年 11 月 13 日
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) );
その他の回答 (1 件)
Jeremy
2019 年 11 月 12 日
編集済み: Jeremy
2019 年 11 月 12 日
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 件のコメント
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