How can i replace 0 by -1 in all binary number till 256 as 00000000, 00000001, 00000010, 00000011, 00000100, ..........​​.........​.​........​..​...., 11111111 ?

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Without using loops

Respuestas (2)

Azzi Abdelmalek
Azzi Abdelmalek el 17 de Jul. de 2015
Editada: Azzi Abdelmalek el 17 de Jul. de 2015
a=dec2bin(0:255)
out=strrep(cellstr(a),'0','1')
  4 comentarios
kailash kumar
kailash kumar el 17 de Jul. de 2015
Sir it turnout to a 256X8 matrix , but all elements are 1. but it should be 0 to 255 binary numbers, in which all 0 should be replaced by '-1'.

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arich82
arich82 el 17 de Jul. de 2015
Editada: arich82 el 17 de Jul. de 2015
[Edited to use 0:255 instead of 1:256]
n = 8;
a = dec2bin(0:n-1) - '0'
returns a as a matrix of doubles:
a =
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
Now just use a logical mask to change 0 --> -1:
a(a == 0) = -1
which returns
a =
-1 -1 -1
-1 -1 1
-1 1 -1
-1 1 1
1 -1 -1
1 -1 1
1 1 -1
1 1 1
For your case, use n = 256. The complete solution (I think) is:
n = 256;
a = dec2bin(0:n-1) - '0';
a(a==0) = -1;
disp(a);
Please accept this answer if it helps, or let me know in the comments if it's not what you're looking for.
  1 comentario
arich82
arich82 el 17 de Jul. de 2015
Alternative to dec2bin: bitget.
Unfortunately, it seems bitget can only get either several bits from a scalar, or a single bit from each element of an array. Wrapping it in bsxfun gets around this apparent limitation:
a = bsxfun(@bitget, [0:255].', 8:-1:1);
a(a == 0) = -1;
Not an issue for your use case, but for much larger arrays, I'd expect bitget to be much quicker than all of the character conversions going on in the dec2bin version, though I could be wrong. (You'd also want to use nextpow2 to get all digits for the more general case).
n = 256;
a = bsxfun(@bitget, (0:n - 1).', nextpow2(n - 1):-1:1);
a(a == 0) = -1;
There's also de2bi if you have the appropriate toolbox (Communications System Toolbox?):

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