Thread Subject:

How to find the indexes of repeated values in a rows in a matrix?

Hi all,

I have a matrix like:

     1 41 0 0 0 0 0 0 0
     1 41 3 0 0 0 0 0 0
     1 41 6 0 0 0 0 0 0
     1 41 13 0 0 0 0 0 0
     1 41 3 48 0 0 0 0 0
     1 41 3 51 0 0 0 0 0
     1 41 6 46 0 0 0 0 0
     1 41 6 47 0 0 0 0 0
     1 41 3 48 4 0 0 0 0
     1 41 3 48 4 41 0 0 0
     1 41 3 48 4 48 0 0 0
     1 41 3 48 4 41 1 0 0
     
I want to count the repeated values per rows, so the output would be:
0
0
0
0
0
0
0
0
0
1 (41 showed up two times)
1 (48 showed up two times)
2 ( 1 and 41 showed up two times)

Any clue of how I can do that?

Thanks

P.

On 5/25/2012 8:26 PM, Paulo Guimaraes wrote:
> Hi all,
>
> I have a matrix like:
>
> 1 41 0 0 0 0 0 0 0
> 1 41 3 0 0 0 0 0 0
> 1 41 6 0 0 0 0 0 0
> 1 41 13 0 0 0 0 0 0
> 1 41 3 48 0 0 0 0 0
> 1 41 3 51 0 0 0 0 0
> 1 41 6 46 0 0 0 0 0
> 1 41 6 47 0 0 0 0 0
> 1 41 3 48 4 0 0 0 0
> 1 41 3 48 4 41 0 0 0
> 1 41 3 48 4 48 0 0 0
> 1 41 3 48 4 41 1 0 0
>
> I want to count the repeated values per rows, so the output would be:
> 0
> 0
> 0
> 0
> 0
> 0
> 0
> 0
> 0
> 1 (41 showed up two times)
> 1 (48 showed up two times)
> 2 ( 1 and 41 showed up two times)
>
> Any clue of how I can do that?
>
> Thanks
>
> P.


But 0 is repeated value? isn't zero a value?

--Nasser

On 5/25/2012 8:26 PM, Paulo Guimaraes wrote:
> Hi all,
>
> I have a matrix like:
>
> 1 41 0 0 0 0 0 0 0
> 1 41 3 0 0 0 0 0 0
> 1 41 6 0 0 0 0 0 0
> 1 41 13 0 0 0 0 0 0
> 1 41 3 48 0 0 0 0 0
> 1 41 3 51 0 0 0 0 0
> 1 41 6 46 0 0 0 0 0
> 1 41 6 47 0 0 0 0 0
> 1 41 3 48 4 0 0 0 0
> 1 41 3 48 4 41 0 0 0
> 1 41 3 48 4 48 0 0 0
> 1 41 3 48 4 41 1 0 0
>
> I want to count the repeated values per rows, so the output would be:
> 0
> 0
> 0
> 0
> 0
> 0
> 0
> 0
> 0
> 1 (41 showed up two times)
> 1 (48 showed up two times)
> 2 ( 1 and 41 showed up two times)
>
> Any clue of how I can do that?
>
> Thanks
>
> P.


This assumes 0 is not part of the logic for
repeated as you show above. Hence I replaced
0 by nan to get this to work:

---------------------------------

B = A;
B(A==0) = nan;
r = zeros(size(B,1),1);

for i = 1:length(r)

      u = unique(B(i,:));
      n = hist(B(i,:),u);
      r(i) = length(n(n>1));

end
----------------------------------

EDU>> r

r =

      0
      0
      0
      0
      0
      0
      0
      0
      0
      1
      1
      2


--Nasser
     

Thanks!

Yes, it was just for nonzero values. Thank you very much.




"Paulo Guimaraes" wrote in message <jppbfu$8fa$1@newscl01ah.mathworks.com>...
> Hi all,
>
> I have a matrix like:
>
> 1 41 0 0 0 0 0 0 0
> 1 41 3 0 0 0 0 0 0
> 1 41 6 0 0 0 0 0 0
> 1 41 13 0 0 0 0 0 0
> 1 41 3 48 0 0 0 0 0
> 1 41 3 51 0 0 0 0 0
> 1 41 6 46 0 0 0 0 0
> 1 41 6 47 0 0 0 0 0
> 1 41 3 48 4 0 0 0 0
> 1 41 3 48 4 41 0 0 0
> 1 41 3 48 4 48 0 0 0
> 1 41 3 48 4 41 1 0 0
>
> I want to count the repeated values per rows, so the output would be:
> 0
> 0
> 0
> 0
> 0
> 0
> 0
> 0
> 0
> 1 (41 showed up two times)
> 1 (48 showed up two times)
> 2 ( 1 and 41 showed up two times)
>
> Any clue of how I can do that?
>
> Thanks
>
> P.

"Paulo Guimaraes" wrote in message <jppbfu$8fa$1@newscl01ah.mathworks.com>...
> Hi all,
>
> I have a matrix like:
>
> 1 41 0 0 0 0 0 0 0
> 1 41 3 0 0 0 0 0 0
> 1 41 6 0 0 0 0 0 0
> 1 41 13 0 0 0 0 0 0
> 1 41 3 48 0 0 0 0 0
> 1 41 3 51 0 0 0 0 0
> 1 41 6 46 0 0 0 0 0
> 1 41 6 47 0 0 0 0 0
> 1 41 3 48 4 0 0 0 0
> 1 41 3 48 4 41 0 0 0
> 1 41 3 48 4 48 0 0 0
> 1 41 3 48 4 41 1 0 0
>
> I want to count the repeated values per rows, so the output would be:
> 0
> 0
> 0
> 0
> 0
> 0
> 0
> 0
> 0
> 1 (41 showed up two times)
> 1 (48 showed up two times)
> 2 ( 1 and 41 showed up two times)
>
> Any clue of how I can do that?

Assuming "valid" elements of A are > 0:

sum(diff(diff([false(size(A,1),1) sort(A,2)],1,2)<=0,1,2)==1,2)

% Bruno