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How to simulate bit stuffing on a 20860×1 matrix of binary data?

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Hello everyone,
i have a matrix of binary data of size 20860×1. (20860 rows and 1 column), and i would like to perform bit stuffing on it, which means to insert a zero after each six consecutive ones
so if i have a sequence like this in my matrix: 0 0 0 1 1 1 1 1 1 1 0
then after the bit stuffing it will be like this: 0 0 0 1 1 1 1 1 1 0 1 0
i tried doing this using two for loops but it didn't work properly
let the data matrix be called: collect, and let it be a matrix of random binary data
collect=randi([1 0],20860,1);
  3 comentarios
Rania Fahmy
Rania Fahmy el 14 de Dic. de 2018
Editada: Rania Fahmy el 14 de Dic. de 2018
they do have to be consecutive yes
Fangjun Jiang
Fangjun Jiang el 14 de Dic. de 2018
I can suggest utilizing sparse matrix, see sparse().

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Respuesta aceptada

Guillaume
Guillaume el 14 de Dic. de 2018
An easy way:
%demo array that will fail with most naive algorithms:
v = [ones(1, 8), 0, ones(1, 5), 0, ones(1, 20)]
newv = regexprep(char(v + '0'), '111111', '1111110') - '0'

Más respuestas (1)

Omer Yasin Birey
Omer Yasin Birey el 14 de Dic. de 2018
Hi Rania, you may try this:
a =randi([0 1],1,20860);
a = num2str(a);
a = strsplit(a);
a = strcat(a);
a = cell2mat(a);
k = strfind(a,'111111');
for i = 1:length(k)
b = [a(1:k(i)+6) '0' a(k(i)+8:end)];
end
b = b';
  2 comentarios
Guillaume
Guillaume el 14 de Dic. de 2018
Editada: Guillaume el 14 de Dic. de 2018
I've not tried to fully understand what this answer is doing. I'll note that conversion to string (and back) is going to be slow.
At the very least, for a vector of 0 and 1, a conversion to string with:
a = randi([0 1], 1, 20860);
str = char(a + '0');
is going to be much faster than num2str.
Note that although undocumented strfind also works with vectors of numbers.
Also, I suspect that this answer will insert a 0 after the 6th, 7th and 8th one in a sequence of 8 consecutive 1.
edit: actually, it would do if the loop wasn't completely broken. As it is, it just insert a 0 at the end of the last sequence no matter how many there are.
Rania Fahmy
Rania Fahmy el 14 de Dic. de 2018
I just tried this, and it gives me the same original matrix. For some reason it just didn't add any zeros after the occurance of 6 consecutive ones.

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