Finding the previous consecutive entries in a matrix column

I have a matrix A, where upon each iteration, a new row of data is added

A = [1 0 0 1 0; 1 1 0 1 1; 0 1 0 0 1; 0 1 0 1 0; 0 1 0 1 0]; % just a matrix for example

Each 1 in each row A(i) represents a species pairing to another species, 1 in A(i,j), where j represents a nonspecific column, and pairing happens because a certain condition is met. I have this:

%Will not produce A above, just an example of my current code
iterations = 5; %random numbers for example sake
number_of_species = 10;

original_conditions = [1 0 0 1 0];
A=[original_conditions;zeros(iterations,number_of_species)]];

for j=1:number_of_species
    for k = j+1:number_of_species
            if condition   %placeholder, my condition would take several lines of code
                A(i,j) = 1;
                A(i,k) = 1; 
end

I need these 1's to stay paired for a certain amount of time, time_paired. Each iteration represents the change in time, dt.

So, for example, species_p = A(2,2) is paired on the second iteration. I need to keep it paired until it has been paired long enough to satisfy time_paired. After some research, I think the continue command is best for not affecting species_p until it has satisfied time_paired. But I have not been able to identify how to get a count for identifying the immediate consecutive entries in the rows, within the same column, above my current species. I do not need a count of all of the consecutive entries above my species, just the ones immediately above it. This is my main issue, and I think i could figure out the following question if I were able to figure this out.

For simplicity, let dt = 1, time paired = 2, iterations = 2: dt*iterations=time_paired.

Upon the 3rd iteration, species_p = A(3,2) has satisfied time_paired. This is not a problem because species_p would be able to be processed through the rest of my program. However, on the 4th iteration, species_p has been paired with a different species than it was in the 3rd iteration, but as my logic stands, it would be released even though it should be held for another iteration. I figured when the number of previous consecutive entries in the column (dt*iterations) exceeds time_paired, i could subtract iterations=time_paired/dt from this.

I have tried several methods proposed in answers to questions like this, but none of them fit.

[Jan]

The formulation of the question is complicated. You have this input:

A = [1 0 0 1 0; ...
     1 1 0 1 1; ...
     0 1 0 0 1; ...
     0 1 0 1 0; ...
     0 1 0 1 0];

What is the wanted output? Should every 1 be continures by further 1s on the right side?

[Stephen23]

apple3's incorrectly posted "Answer" moved here:

A is produced after 4 iterations, but I basically need a function that will tell me, at A(4,2), there are two consecutive ones directly above it. I can trouble shoot the rest.

A more general form of telling me how many 1’s are consecutively above A(i,j) would be awesome.

Sorry if I phrased this bad. I think I was trying to ask too many questions at once.

[Jan]

Programming means to get a simple abstract representation of the problem. So for the given example input:

A = [1 0 0 1 0; ...
     1 1 0 1 1; ...
     0 1 0 0 1; ...
     0 1 0 1 0; ...
     0 1 0 1 0];

the wanted output is:

B = [0 0 0 0 0; ...
     1 0 0 1 0; ...
     0 1 0 0 1; ...
     0 2 0 0 0; ...
     0 3 0 1 0];

Then:

B = max(cumsum(A, 1) - 1, 0);

Or should B(4,3) be 2 and B(5,3) be 3? This would mean, that the cummulative sum is reset, when the value is 0.

Do you see, that the wanted procedure can be explained much more compact?

[apple3]

Thank you so much! Yes I see how that is more compact. I was thinking about it wrong. Thank you!

Answer 1

[Jan]

AS far as I understand the discussion, this solves your needs:

B = max(cumsum(A, 1) - 1, 0);