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Find the number of unique rows and its corresponding index in a matrix

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Hi there, I need to list the unique rows and its corresponding index in a matrix without the using the unique function. I have been suggested to use the unique function but it seems rather time consuming and I have to run many iterations. Any help will be appreciated. Cheers, Vishal
Data:
v1=[1 2 3 4 5; 5 3 3 2 1; 1 2 4 9 7; 1 2 3 4 5; 5 3 3 2 1]
[a,b,c]=unique(v1,'rows','stable')
The required output which I get from the unique function is:
a =
1 2 3 4 5
5 3 3 2 1
1 2 4 9 7
b =
1
2
3
c =
1
2
3
1
2
  7 comentarios
dpb
dpb el 27 de Ag. de 2013
Editada: dpb el 27 de Ag. de 2013
Well, it's a big job from scratch every time.
Is this simulation the selection of a subset each time that I seem to recall something about in an earlier posting? If so, I'd suggest the solution may be to do it once for the complete set and the use that and select from it for each randomized sample as well as from the data. That is if the data itself aren't changing, only the sample space.
Jing
Jing el 28 de Ag. de 2013
How long does it take? I don't think UNIQUE will take a lot of time...and my test is running 20 times the UNIQUE line with your v1, the running time is always less than 0.005 seconds.

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Jan
Jan el 28 de Ag. de 2013
Editada: Jan el 28 de Ag. de 2013
Perhaps this is more efficient:
v1 = [1 2 3 4 5; 5 3 3 2 1; 1 2 4 9 7; 1 2 3 4 5; 5 3 3 2 1];
vX = v1 * [1, 11, 121, 1331, 14641];
[dummy, b, c] = unique(vX, 'stable');
a = v1(b, :);
unique() has a remarkable overhead, so you could create a local copy of it and remove the not required checks of the inputs etc.
  1 comentario
Vishal
Vishal el 29 de Ag. de 2013
Cheers mate. Shaved off 5% of the computing time. Small typo in line 2
v1 = [1 2 3 4 5; 5 3 3 2 1; 1 2 4 9 7; 1 2 3 4 5; 5 3 3 2 1]; vX = v1 * [1; 11; 121; 1331; 14641]; [dummy, b, c] = unique(vX, 'stable'); a = v1(b, :);

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