Convert Integer Array to Binary Array
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Igor Felipe Gallon
el 15 de Jul. de 2015
Comentada: dpb
el 17 de Mayo de 2019
Hello, people.
I want to convert an Integer array to a Binary array but i don't know how to do this. I've read about the function "dec2bin" but it doesn't do what i exactly want (i think...). For example (sorry my bad artistic skills):

So, thank you!
2 comentarios
Image Analyst
el 15 de Jul. de 2015
Editada: Image Analyst
el 15 de Jul. de 2015
Seems weird to me that all numbers don't have the same number of characters. So the binary number tells you nothing about what created it. For example 1011 could come from both 23 and from 51. What good is that? Why do you want to do this?
Respuesta aceptada
Andrei Bobrov
el 16 de Jul. de 2015
Editada: Andrei Bobrov
el 16 de Jul. de 2015
intArray -> (bitArray, nBit)
intArray = [3 2 1 7 6 2 5];
[~,nBit] = log2(intArray);
bitArray = cell2mat(arrayfun(@dec2bin,V,'un',0));
(bitArray, nBit) -> intArray
intArray = cellfun(@bin2dec,mat2cell(bitArray,1,nBit));
3 comentarios
Nehemias Rodriguez
el 17 de Mayo de 2019
Editada: Nehemias Rodriguez
el 17 de Mayo de 2019
it says undefined V. help please. my array is A = zeros(1, 50);
dpb
el 17 de Mayo de 2019
That was a typo/transcription error from Andrei's desktop to the forum Answer -- he obviously had an array V he used to test and missed getting all references changed to the OPs variable when pasted code into Answer.
V is intArray for the example...
Más respuestas (2)
dpb
el 15 de Jul. de 2015
Using builtins w/o mod/rem to separate digits...
>> dec2bin(str2num(sprintf('%d',x).'))
ans =
011
010
001
111
110
010
101
>>
4 comentarios
Walter Roberson
el 16 de Jul. de 2015
For GA work you will very likely want fixed-length segments. With the scheme you are proposing, after you do the crossover, you do not know how to deconstruct it into integers again, other than the fact that you know that in this scheme no integer will start with a binary 0.
Walter Roberson
el 16 de Jul. de 2015
V = 3217625;
d = sprintf('%d', V) - '0';
b = dec2bin(d) .' - '0';
one_seen = cumsum(b,1) > 0;
bits = b(one_seen) .';
2 comentarios
Walter Roberson
el 16 de Jul. de 2015
You are relying upon accidents of your encoding scheme. If you were to renumber the entries in an array then you would change which mutated states are possible. For example if you were to initialize 20 different array indices all to 1, then those would encode as 1 bit each, and your mutation cannot create entries that represent 2 or higher because additional bits would be required to store those indices.
If your swapping can swap in the "middle" of a symbol then you could end up swapping a 0 from the end of 1 symbol to become the beginning of another symbol, but then you would not be able to decode that. For example, 10|1 with the 2nd and 3rd bits swapped would be 11|0 but 0 is not a valid encoding for any number in your scheme.
If your swapping is restricted to symbol boundaries then you have not gained anything other than complexity of representation compared to using a fixed number of bits to represent the indices.
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