checking for dependency in the previous row?

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abhisrisai on 10 Jul 2019
Commented: abhisrisai on 11 Jul 2019
Hi,
My doubt is regarding accessing the previous 'n' rows, but the catch is, that 'n' is not constant or predefined.
For example :
A = [1 2 3 4; 5 6 7 8; 1 2 3 4; 1 2 3 4; 9 10 11 12]
I want to check for how many previous 'n' rows am I obtaining the same value continiously. In the above example, I know that for the 4th row, I'm getting the same number pattern in the third row. But later from my simulations which contains more than 5000 rows, I might obtain a same value for more than 10 rows continously and a few rows later, might see the same pattern but this time for 6 rows. I have to manually count them everytime.
How can I do this through matlab?
Thank you for the help.
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abhisrisai on 10 Jul 2019
Edited: abhisrisai on 10 Jul 2019
Thanks for the immediate response.
Let's say im interested in the sequence 1 2 3 4 which has occurred atleast twice (sorry I didnt mention that in question).
From your matrix, i can see that rows 3 and 4 have repeated. similarly, rows 6-9 have also repeated.
So the output im expecting is an extra column in the end, which indicates the number of times the sequence has appeared continously before. so for your matrix, row 4 will have a value 2 in the output column and similarly row 9 will have a value of 4 in the output column.
Thank you.
Adam Danz on 10 Jul 2019

Adam Danz on 10 Jul 2019
Edited: Adam Danz on 10 Jul 2019
Here's a demo that counts the consecutive rows of "data" (a matrix) that match "key" (a row vector). The variable "consecutiveCounter" is the output and is concatenated to the end of the data matrix.
% Produce demo data
data = [
1 2 3 4
5 6 7 8
1 2 3 4
1 2 3 4
9 10 11 12
1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4
9 10 11 12
9 10 11 12
0 0 0 0];
key = [1,2,3,4];
keyMatch = all(data == key,2);
dMatch = diff([false; keyMatch==1; false]);
S1 = find(dMatch == 1); % start indices of these sequences
S2 = find(dMatch == -1); % end indices
idxCell = arrayfun(@(x1,x2)((x1:x2-1)-x1+1)',S1,S2,'UniformOutput',false);
consecutiveCounter = zeros(size(keyMatch));
consecutiveCounter(keyMatch==1) = cell2mat(idxCell);
% If you want to add that as a 5th column,
dataNew = [data, consecutiveCounter];
Result
dataNew =
1 2 3 4 1
5 6 7 8 0
1 2 3 4 1
1 2 3 4 2
9 10 11 12 0
1 2 3 4 1
1 2 3 4 2
1 2 3 4 3
1 2 3 4 4
9 10 11 12 0
9 10 11 12 0
0 0 0 0 0
abhisrisai on 11 Jul 2019
Thank you very much for the help! It's as expected. :)