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 comentarios
Stephen23
el 27 de Jul. de 2020
"...since the "first" option is not supported by uniquetol? "
According to the documentation, uniquetol returns the index for the first row of each set of matching rows: "Index to A, returned as a column vector of indices to the first occurrence of repeated elements".
It is easy to check that it returns the first index of multiple matching rows:
>> M = [0.0001,2;0.0002,2;1e2,2]
M =
0.0001 2.0000
0.0002 2.0000
100.0000 2.0000
>> [U,X] = uniquetol(M,1e-3,'ByRows',true)
U =
0.0001 2.0000
100.0000 2.0000
X =
1
3
Given that uniquetol returns the index of first occurence by default and apparently that is what you want, what exactly is the problem? Can you given an example where this does not return what you need?
Bruno Luong
el 27 de Jul. de 2020
This doc is obviously not correct or not precise, seems only applicable for EXACT match. If you run my code bellow where tolerance occurs you'll see the 'first' is not fullfilled.
Stephen23
el 27 de Jul. de 2020
Editada: Bruno Luong
el 27 de Jul. de 2020
"This doc is obviously not correct or not precise, seems only applicable for EXACT match."
Did you look at my example? The first two rows are NOT an exact match.
Bruno Luong
el 27 de Jul. de 2020
Editada: Bruno Luong
el 27 de Jul. de 2020
Oops I edit out your post! (now I restore it)
Your example is too small, too specific. When it works for one example doesn't mean it works for all cases.
Stephen23
el 27 de Jul. de 2020
"Your example is too small, too specific. When it works for one example doesn't mean it works for all cases."
A small repeatable example would be useful. No random numbers.
Bruno Luong
el 27 de Jul. de 2020
Editada: Bruno Luong
el 27 de Jul. de 2020
Why? if it fails once (on random numbers) then it simply fails, but anyway I'll prepare you a deterministic example.
Here you can try my random example
A = ceil(3*rand(1000,2));
A = A + rand(size(A))*1e-10;
[Au,matlabidx] = uniquetol(A,1e-6,'byrows',true);
[tf,I] = ismembertol(A,Au,1e-6,'byrows',true);
firstidx = accumarray(I,(1:size(A,1))',[],@min);
% Check index returned
matlabidx
firstidx
norm(A(matlabidx,:)-A(firstidx,:),'fro') % small if they match
I get this (R2020a, Update4, Windows), it shows clearly MATLAB does not return smallest indexes
matlabidx =
868
524
502
489
810
97
537
91
308
firstidx =
2
12
4
1
6
40
5
13
9
ans =
1.9880e-10
Bruno Luong
el 27 de Jul. de 2020
Editada: Bruno Luong
el 27 de Jul. de 2020
Here we go, small example
>> A = 1 + eye(2)*2^-40
A =
1.0000 1.0000
1.0000 1.0000
>> [Au,matlabidx] = uniquetol(A,1e-6,'byrows',true)
Au =
1.0000 1.0000
matlabidx =
2
Bruno Luong
el 27 de Jul. de 2020
Editada: Bruno Luong
el 27 de Jul. de 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
J AI
el 27 de Jul. de 2020
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