How to replace arrays containing consecutive repeated numbers by zero?

I have a nested cell array that consists of vectors as shown in Fig. below:

I want to go through each cell array and check it if contains atleast 50% of consecutive repetated numbers. If the cell array contains such numbers, then I want to fill the entire cell array with zeros.

See example below. In this case all the elements of cell array contains consecutive repeated numbers. I want to replace the entire cell array with zeros.

Note: The reason why I mentioned 50% is that, there were many cell arrays where the first 36,000 or so elements were normal but the other half (36,000) contained consecutive repeated numbers.

Answer 1

[Adam Danz]

For some reason I'm drawn to detecting / counting consecutive numbers. I must have answered more than 5 questions similar to this one and every time I come up with a different convoluted solution that is usually way too difficult to read without dissecting it. This one's no exception. Let's see if it works with your data.

% Demo data
out = {[5;5;5;1;2;3];[1;2;2;2;5;5;5];[1;5;5;5;1];[0;1;1;0;1;1]}; 

%     out{1} =    out{2} =     out3{3} =    out{4} = 
%          5           1             1           0
%          5           2             5           1
%          5           2             5           1
%          1           2             5           0
%          2           5             1           1
%          3           5             
%                      5

% Compute the length of the largest consecutive segment divided by
% the length of the full vector.  
p = cellfun(@(v)max(accumarray(cumsum(diff(v([2,1:end]))~=0)+1,1))/numel(v), out);

% p =
%           0.5
%       0.42857
%           0.6
%       0.33333

% Replace entire vectors in "out" that contain a segment of consecutive values 
% equal to or longer than 50% of the length of the vector, with zeros. 
out(p>=0.5) = cellfun(@(v){zeros(size(v))},out(p>=0.5))

%     out{1} =    out{2} =     out3{3} =    out{4} = 
%          0           1             0           0
%          0           2             0           1
%          0           2             0           1
%          0           2             0           0
%          0           5             0           1
%          0           5             
%                      5

The much-easier-to-read, unpacked, loop version of the cellfun() would look like

p = nan(size(out)); 
for i = 1:numel(out)
    diffOut = diff(out{i}([2,1:end]));    % 0s indicate a consecutive value.
    nonConsecCount = cumsum(diffOut~=0);  % Group consec. vals. with 0s and non-consec-vals with cumulative integers
    consecCount = accumarray(nonConsecCount+1, 1); % The number of consec values for each segment
    p(i) = max(consecCount)/numel(out{i}); % largest number of consec. /  length of vector
end

[maruljay]

Thank you so much. It worked like a charm. You just saved me tonnes of hours of work.

Your code explanation is crystal clear.

[Adam Danz]

Glad it worked, thanks for letting us know!

Answer 2

[Image Analyst]

Use a for loop and diff(). If you can't figure it out, attach a small chunk of your cell array in a .mat file.

[maruljay]

Thanks for the insight. I have accepted the answer given by Adam Danz. It worked.