Better formulation of looking for duplicate numbers

I am sorry for my poor English - A matrix of different positive
integers should be a partial result of my program. The program always
produces in each cycle a unique matrix as a result of some
complicated calculations. But sometimes things goes wrong and some
numbers appear twice in the matrix. That should not happen. I want to
find these duplicate numbers and refine the calculation for these
numbers - so my point is not to generate an arbitrary matrix of
different positive integers but to find mistakes in my resulting
model. thanks alot jakub

 

Hi Jakub.. how about something like..

mat_size=size(yourmatrix);
for i=1:mat_size(1,1)
for j=1:mat_size(1,2)
locs=find(yourmatrix==matrix(i,j));

if(length(locs)>1)
% dO THE repeat calculation for locs
end
end
end

i am sure there is a more elegant way of doing this...but this might
help.

Ashu

 

Results=magic(3);
Results=cat( 1, Results, Results( 1, )
[b,m,n]=unique(Results);
Results(m)=[] % prints the copies

/ per

 

Hi! Thanks. Well this is a way how to get around but it's not good.
It's better to avoid the sequential algorithms like this "for-end". I
am working with huge chunks of data and scanning through the whole
matrix would take too long. I was wondering if there was some exact
command to deal with this or some ingenious trick someone has
invented. I will try to think hard to come up with one and post it
here.
Anyway thanks for your time... jakub

 

Try using hist.

Consider:

A = floor(rand(10)*100);
[n,x] = hist(A),unique(A)));

The n's will tell you the number of times each x occurs in A. I leave it to
you to find the repeated values in your A matrix.

Cheers,
Brett

 

The reason - why posted this was:
1. My ignorance to "unique"
2. I feel that now for loops do run as fast as vectorized code and
that said - if you do have a large matrix - sooner or later i imagine
one has to bite the bullet in terms of comp. time.

The way - look at it is - no matter how it's done - physically the
process has to sieve through all elements once to look for any
duplicacies..

That's just my rather limited take..

Regards,

Ashu

 

Per: Brett:

Thanks a lot - You have the right ways of doing this. I am happy.

regards.