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

Asked by aamir irshad on 27 Jun 2018

Latest activity Commented on by aamir irshad on 27 Jun 2018

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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

2 Comments

Paolo on 27 Jun 2018

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

aamir irshad on 27 Jun 2018

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.

3 Answers

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Answer by Stephen Cobeldick on 27 Jun 2018

Accepted Answer

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

1 Comment

aamir irshad on 27 Jun 2018

Hi Stephen,

This is most efficient. Thanks for your help.

BR,

Aamir

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Answer by Jan on 27 Jun 2018

Edited by Jan on 27 Jun 2018

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.

1 Comment

aamir irshad on 27 Jun 2018

Its also working perfectly.

Thank you, Jan for another efficient logic.

BR,

Aamir

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Answer by Ameer Hamza on 27 Jun 2018

Edited by Ameer Hamza on 27 Jun 2018

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

5 Comments

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aamir irshad on 27 Jun 2018

Great.

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

Thanks a lot.

Aamir

Ameer Hamza on 27 Jun 2018

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 on 27 Jun 2018

Thanks a lot

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