Find repeating rows in a matrix and replace with zeros

I have a large matrix (5683384x2 double) containing xy-coordinates of an ROI mask. If an xy-coordinate does not repeat enough times, I want to set the row to zeros.

For example if my matrix is [9 10; 9 10; 9 10; 9 10; 9 10; 7 12; 3 4; 3 4; 3 4; 3 4]. I want to replace rows that do not repeat 3 times, so [7 12] would become [0 0]. Is there any way to do this?

Answer 1

[Monika Jaskolka]

This can probably be optimized, but it works to replace rows that don't occur more times than the set threshold, in this case 3 times.

M = [9 10; 9 10; 9 10; 9 10; 9 10; 7 12; 3 4; 3 4; 3 4; 3 4; 7 12];
threshold = 3;
 
[~, ~, ic] = unique(M, 'rows', 'stable');
numOccurances = accumarray(ic, 1);
idxToDelete = false(size(M,1), 1);
for i = 1:length(numOccurances)
    if numOccurances(i) < threshold        
        duplicateRows = find(ic==i);
        idxToDelete(duplicateRows) = true;
    end
end
M(idxToDelete,:) = 0;

If you want to delete the rows entirely, you can replace the last line with this:

M(idxToDelete,:) = []; % Alternatively, delete rows

[Savannah D]

Thank you! the numOccurances tripped me up at first with my threshold, but this worked great!

[Monika Jaskolka]

No problem! If the answer worked for you, please press the "Accept this answer" button.

Answer 2

[Bruno Luong]

M = [9 10; 9 10; 9 10; 9 10; 9 10; 7 12; 3 4; 3 4; 3 4; 3 4; 7 12];
minlgt = 3; % keep all rows consecutively repeated at least minlgt times

i = find([true; any(diff(M,1,1),2); true]);
lgt = diff(i);
removed = lgt<minlgt;
M(repelem(i(removed),lgt(removed)),:) = 0