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From: Jack on 28 Jun 2010 19:06
Hi,

I have a matrix consisting of a large number of rows. I need to know which rows has been repeated exactly the same more than once and how many times they have been repeated.
For Example if A= [ 1 2 3 ; 4 3 5; 1 2 3; 1 2 3; 4 3 5; 5 2 1; 3 2 1; 3 5 1]

then I should get something like the following:
1 2 3 : 3 times
4 3 5: 2 times

I wonder if there is an easy way to do it?
many thanks
From: Matt Fig on 28 Jun 2010 19:33
How about a hard way?

% Data
A= [ 1 2 3 ; 4 3 5; 1 2 3; 1 2 3; 4 3 5; 5 2 1; 3 2 1; 3 5 1]


% Engine
B = sortrows(A);
S = [1;any(diff(B),2)];
[L,S] = regexp(sprintf('%i',S'),'1(0)+','start','end');
repeated_rows = B(S,:) % Repeated Rows.
repeat_count = (S-L+1)' % How often each repeated row appears.
From: Jack on 30 Jun 2010 18:06
"Matt Fig" <spamanon(a)yahoo.com> wrote in message <i0bbg4$l1k$1(a)fred.mathworks.com>...
> How about a hard way?
>
> % Data
> A= [ 1 2 3 ; 4 3 5; 1 2 3; 1 2 3; 4 3 5; 5 2 1; 3 2 1; 3 5 1]
>
>
> % Engine
> B = sortrows(A);
> S = [1;any(diff(B),2)];
> [L,S] = regexp(sprintf('%i',S'),'1(0)+','start','end');
> repeated_rows = B(S,:) % Repeated Rows.
> repeat_count = (S-L+1)' % How often each repeated row appears.

My Dear;

Thanks alot for your response; I found it by coincedence. I did not receive an email "as usual" letting me know someone has responded... this is weired. I will go and check the code right now.

Many thanks
From: sscnekro on 30 Jun 2010 21:13
> Thanks alot for your response; I found it by coincedence. I did not receive an email "as usual" letting me know someone has responded... this is weired

Now I start to understand why some guys don't know about the help they really receive. They wait for e-mail instead of coming to the page. And it's also true that there were two or three cases when I e-mailed an alert on my nwsgr. reply (when the person was a first poster and seemed too desperate).
From: Jack on 1 Jul 2010 00:33
"Matt Fig" <spamanon(a)yahoo.com> wrote in message <i0bbg4$l1k$1(a)fred.mathworks.com>...
> How about a hard way?
>
> % Data
> A= [ 1 2 3 ; 4 3 5; 1 2 3; 1 2 3; 4 3 5; 5 2 1; 3 2 1; 3 5 1]
>
>
> % Engine
> B = sortrows(A);
> S = [1;any(diff(B),2)];
> [L,S] = regexp(sprintf('%i',S'),'1(0)+','start','end');
> repeated_rows = B(S,:) % Repeated Rows.
> repeat_count = (S-L+1)' % How often each repeated row appears.

Dear Matt;
Thanks a lot, this works well :) but it is really a hard way at least for me, I had to go to the help menu many times!! but how could I make change to the code if I want to list all the rows ( not only those repeated more than once)?
Best Regards
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