How to pad extra 0 after 5 consecutive 1's in an array

I need to insert extra 0 in a large array after consecutive 5 ones e.g, if

 a=[0 1 1 1 1 1 0 1 1 1 1 1 1 0 0 1 1 1 1 1 0]..........so on
                 output array should be

b=[0 1 1 1 1 1 0 0 1 1 1 1 1 0 1 0 0 1 1 1 1 1 0 0 ]..........so on

Thanks in advance

[Paolo]

In your input there is a sequence of 6 1s which become 5 1s. Is that also a requirement?

[aamir irshad]

thanks for your reply. Yes, I can explain it a bit more e.g. if we have a= [1 1 1 1 1 1 1 1 1 1 1 1] the output should be b= [1 1 1 1 1 0 1 1 1 1 1 0 1 1] The logic should insert an extra 0 after consecutive 5 ones start looking from the beginning of data.

Answer 1

[Stephen Cobeldick]

Simpler:

>> a = [0 1 1 1 1 1 0 1 1 1 1 1 1 0 0 1 1 1 1 1 0];
>> b = regexprep(char(a+'0'),'11111','111110')-'0'
b =
   0   1   1   1   1   1   0   0   1   1   1   1   1   0   1   0   0   1   1   1   1   1   0   0
>> c = [0 1 1 1 1 1 0 0 1 1 1 1 1 0 1 0 0 1 1 1 1 1 0 0 ] % your requested output
c =
   0   1   1   1   1   1   0   0   1   1   1   1   1   0   1   0   0   1   1   1   1   1   0   0

[aamir irshad]

Hi Stephen,

This is most efficient. Thanks for your help.

BR,

Aamir

Answer 2

[Jan]

a = [1 1 1 1 1 1 1 1 1 1 1 1];

b = a;
idx = strfind(a, [1,1,1,1,1])
if ~isempty(idx)
  % Remove indices with a too short distance:
  p = idx(1);
  for k = 2:numel(idx)
    if idx(k) - p < 5
       idx(k) = 0;
    else
       p = idx(k);
    end
  end
  idx = idx(idx ~= 0) + 4;
  %
  b         = cat(1, b, nan(size(b)));  % Pad with NaNs
  b(2, idx) = 0;                        % Insert zeros
  b         = b(~isnan(b)).';           % Crop remaining NaNs
end

This avoids changing the size of the output array in each iteration, which should be more efficient for large data sets.

[aamir irshad]

Its also working perfectly.

Thank you, Jan for another efficient logic.

BR,

Aamir

Answer 3

[Ameer Hamza]

Try this

a= [1 1 1 1 1 1 1 1 1 1 1 1];
b = a;
count = 1;
while true
  index = strfind(b(count:end), [1 1 1 1 1])+5;
  if isempty(index)
      break
  end
  index = index(1);
  count = count+index-1;
  b = [b(1:count-1) 0 b(count:end)];
end
b

b =
Columns 1 through 13
   1     1     1     1     1     0     1     1     1     1     1     0     1
Column 14
   1

[aamir irshad]

Hi, Thanks for your reply. It is not working as desired because my data is not same all the time. Please see my reply to Paolo above. Starting from the first element the logic should insert extra 0 after 5 consecutive ones.

[Ameer Hamza]

Sorry! My bad. I wrongfully assumed that the 5 ones will always be preceded by a zero. I have fixed it, try the updated code. Try it now.

[aamir irshad]

Great.

It is working perfectly. Could you explain the logic a bit?

Thanks a lot.

Aamir

[Ameer Hamza]

The code finds the first occurrence of [1 1 1 1 1] in the array and add a zero after that. In the next iteration, it skips the first part of the array and searches the remaining array. If another occurrence of 5 1s is found, it will again add zero and search the remaining array. It will continue until all the groups of 5 1s have passed.

You can set a breakpoint at first line of the code and execute the each line one by one to better understand the logic.

[aamir irshad]

Thanks a lot