How to pad extra 0 after 5 consecutive 1's in an array
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aamir irshad
il 27 Giu 2018
Commentato: aamir irshad
il 27 Giu 2018
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 Commenti
Paolo
il 27 Giu 2018
In your input there is a sequence of 6 1s which become 5 1s. Is that also a requirement?
Risposta accettata
Stephen23
il 27 Giu 2018
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
Più risposte (2)
Jan
il 27 Giu 2018
Modificato: Jan
il 27 Giu 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.
Ameer Hamza
il 27 Giu 2018
Modificato: Ameer Hamza
il 27 Giu 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 Commenti
Ameer Hamza
il 27 Giu 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.
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