How to delete ones in a binary vector that correspond to more than k consecutive repetitions?

Given a number k and a binary vector v. How to saturate the number of consecutive repetitions of ones in v to the value k? For example, if k=3 and v=[1,0,0,1,1,0,1,1,1,1,0,0] then the output should be w=[1,0,0,1,1,0,1,1,1,0,0]. Speed of the code is highly important and no toolboxes available.

Thanks in advance

Answer 1

[Ameer Hamza]

Try this. It uses function from image processing toolbox. You can check if the speed is acceptable

v = [1,0,0,1,1,0,1,1,1,1,0,0,1,1,1,1,1,0,1];

rgs = bwconncomp(v);
pxlList = cellfun(@(x) {(x(4:end).')}, rgs.PixelIdxList);
v([pxlList{:}]) = [];

Result:

v =
  Columns 1 through 13
     1     0     0     1     1     0     1     1     1     0     0     1     1
  Columns 14 through 16
     1     0     1

[GONZALEZ DE COSSIO ECHEVERRIA Francisco Jose]

Thank you for your answer. However, I was wondering if there is a simpler way without the need of external toolboxes.

[Ameer Hamza]

Try this code, it does not require any external toolbox.

v = [1,1,1,1,0,0,1,1,0,1,1,1,1,0,0,1,1,1,1,1,0,1];

zloc = [0 find(v==0)];
dzloc = find(diff(zloc) > 3);
idx = cell2mat(arrayfun(@(x) {zloc(x)+4:zloc(x+1)-1}, dzloc));
v(idx) = [];

[GONZALEZ DE COSSIO ECHEVERRIA Francisco Jose]

Thanks a lot, it works fine.

[Ameer Hamza]

I am glad to be of help.