Replace repeated value based on sequence size

I have a 2D matrix composed of ones and zeros.

mat = [0 0 0 0 1 1 1 0 0
       1 1 1 1 1 0 0 1 0
       0 0 1 0 1 1 0 0 1];

I need to find all consecutive repetitions of ones in each row and replace all ones with zeros only when the sequence size is smaller than 5 (5 consecutive ones):

mat = [0 0 0 0 0 0 0 0 0
       1 1 1 1 1 0 0 0 0
       0 0 0 0 0 0 0 0 0];

Any suggestion on how to approach this problem would be very welcome.

CodePudding user response：

You can use diff to find the start and end points of the runs of 1, and some logic based on that to zero out the runs which are too short. Please see the below code with associated comments

% Input matrix of 0s and 1s
mat = [0 0 0 0 1 1 1 0 0
       1 1 1 1 1 0 0 1 0
       0 0 1 0 1 1 0 0 1];
% Minimum run length of 1s to keep   
N = 5;

% Get the start and end points of the runs of 1. Add in values from the
% original matrix to ensure that start and end points are always paired
d = [mat(:,1),diff(mat,1,2),-mat(:,end)];
% Find those start and end points. Use the transpose during the find to
% flip rows/cols and search row-wise relative to input matrix.
[cs,r] = find(d.'>0.5);  % Start points
[ce,~] = find(d.'<-0.5); % End points
c = [cs, ce];            % Column number array for start/end
idx = diff(c,1,2) < N;   % From column number, check run length vs N

% Loop over the runs which didn't satisfy the threshold and zero them
for ii = find(idx.')
    mat(r(ii),c(ii,1):c(ii,2)-1) = 0;
end

CodePudding user response：

The vectorized solution by @Wolfie is nice and concise, but a bit hard to understand and far from the wording of the problem. Here is a direct translation of the problem using loops. It has the advantage of being easier to understand and is slightly faster with less memory allocations, which means it will work for huge inputs.

[m,n] = size(mat);
for i = 1:m
    j = 1;
    while j <= n
        seqSum = 1;
        if mat(i,j) == 1
            for k = j 1:n
                if mat(i,k) == 1
                    seqSum = seqSum   1;
                else
                    break
                end
            end
            if seqSum < 5
                mat(i,j:j seqSum-1) = 0;
            end
        end
        j = j   seqSum;
    end
end