Finding unique rows using "uniquetol" from top
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It seems like there is no option for finding unique rows from top using uniquetol unlike unique where there is option for the argument "first". Is there a way to do this
[~,idx] = uniquetol(Q2(:,1:ns),'ByRows',true,'first') % the argument "first" picks unique rows from the top
some other way, since the "first" option is not supported by uniquetol?
Thanks!
9 commentaires
Bruno Luong
le 27 Juil 2020
Modifié(e) : Bruno Luong
le 27 Juil 2020
Even smaller (smallest example)
>> A = 1 + [1;0]*2^-40
A =
1.0000
1.0000
>> [Au,matlabidx] = uniquetol(A,1e-6,'byrows',true)
Au =
1
matlabidx =
2
Réponse acceptée
Bruno Luong
le 27 Juil 2020
Modifié(e) : Bruno Luong
le 27 Juil 2020
I suppose you can do something like this. I'm affraid the way UNIQUETOL and ISMEMBERTOL consider the TOL internally, and in some cases the result is not coherently match when the frontier is fuzzy. You might set 'DataScale' to 1 and control TOL to have more robust match.
Fake data
A = ceil(3*rand(1000,2));
A = A + rand(size(A))*1e-10;
Engine
[Au,idx] = uniquetol(A,1e-6,'DataScale',1,'byrows',true);
[tf,I] = ismembertol(A,Au,1e-6,'DataScale',1,'byrows',true);
if ~all(tf)
error('incompatible tolerance');
end
firstidx = accumarray(I,(1:size(A,1))',[],@min)
if ~all(firstidx) || size(firstidx,1)~=size(Au,1)
error('incompatible tolerance');
end
Check
A(idx,:)
A(firstidx,:)
If those matching error checking bother you, here is a way to ignore with the risk that the result might be different than UNIQUETOL alone
Au = uniquetol(A,1e-6,'DataScale',1,'byrows',true);
[tf,I] = ismembertol(A,Au,1e-6,'DataScale',1,'byrows',true);
firstidx = accumarray(I(tf),find(tf),[],@min)
Au = A(firstidx,:);
2 commentaires
Bruno Luong
le 27 Juil 2020
Modifié(e) : Bruno Luong
le 27 Juil 2020
Actually I was stupid; you can use the third output of UNIQUETOL, no need fot ISMEMBERTOL
[Au,matlabidx,I] = uniquetol(A, 1e-6, 'byrows', true);
firstidx = accumarray(I, (1:size(A,1))', [], @min)
% Check
norm(A(matlabidx,:)-A(firstidx,:),Inf)
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