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

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aamir irshad
aamir irshad 2018-6-27
评论： aamir irshad 2018-6-27
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 个评论
Paolo
Paolo 2018-6-27
In your input there is a sequence of 6 1s which become 5 1s. Is that also a requirement?
aamir irshad
aamir irshad 2018-6-27
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.

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采纳的回答

Stephen23
Stephen23 2018-6-27
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

更多回答（2 个）

Jan
Jan 2018-6-27
编辑：Jan 2018-6-27
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.
Ameer Hamza
Ameer Hamza 2018-6-27
编辑：Ameer Hamza 2018-6-27
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 个评论
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Ameer Hamza
Ameer Hamza 2018-6-27
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.

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