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Converting decimal integer to binary vector

On Programmer » Matlab

5,335 words with 8 Comments; publish: Wed, 07 May 2008 10:45:00 GMT; (20093.75, « »)

Hi all

I'm looking for a neat way to produce a vector v with ones and zeros

corresponding to the binary form of a given decimal integer n.

dec2bin by itself is not the answer since that produces a string.

An ugly solution that I know of is

n = 123;

v = dec2bin(n) - 48;

This works since 0 and 1 happens to be next to each other in the

ASCII table, numbered 48 and 49 respectively. Adding or subtracting

to a string converts it to a vector with the ASCII numbers of its

characters.

Now, is there a neat ASCII-independent way to do this?

Thanks in advance

/Fred

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  • 8 Comments
    • Fred wrote:

      >

      > Hi all

      > I'm looking for a neat way to produce a vector v with ones and

      > zeros

      > corresponding to the binary form of a given decimal integer n.

      > dec2bin by itself is not the answer since that produces a string.

      > An ugly solution that I know of is

      > n = 123;

      > v = dec2bin(n) - 48;

      > This works since 0 and 1 happens to be next to each other in the

      > ASCII table, numbered 48 and 49 respectively. Adding or subtracting

      > to a string converts it to a vector with the ASCII numbers of its

      > characters.

      > Now, is there a neat ASCII-independent way to do this?

      > Thanks in advance

      > /Fred

      Hi!

      Try:

      v=dec2bin(n)-'0';

      HTH

      /PB

      #1; Wed, 07 May 2008 10:47:00 GMT
    • PB wrote:

      >

      > Fred wrote:

      string.

      the

      > subtracting

      its

      > Hi!

      > Try:

      > v=dec2bin(n)-'0';

      You could do worse that having a look at TMW's dec2bin function. It

      is an m-file. I guess you could write your own without the char()

      conversion at the end. The meat of the code looks like (d is your

      dec value):

      n=1;

      [f,e]=log2(max(d));

      rem(floor(d*pow2(1-max(n,e):0)),2)

      #2; Wed, 07 May 2008 10:48:00 GMT
    • PB wrote:

      >

      > Fred wrote:

      string.

      the

      > subtracting

      its

      > Hi!

      > Try:

      > v=dec2bin(n)-'0';

      Well, it works. It still depends on '0' and '1' being next to each

      other in the character table though. I was thinking that there were a

      more general preferred way, making use of other convention functions

      or something. I happy with this, but anyone else trying to understand

      such code would be puzzled.

      /Fred

      #3; Wed, 07 May 2008 10:49:00 GMT
    • On Mon, 06 Mar 2006 11:03:17 -0500, Fred wrote:

      > Well, it works. It still depends on '0' and '1' being next to each

      > other in the character table though. I was thinking that there were a

      > more general preferred way, making use of other convention functions

      > or something. I happy with this, but anyone else trying to understand

      > such code would be puzzled.

      How about this?

      v=dec2bin(n)=='1';

      Dan

      > /Fred

      #4; Wed, 07 May 2008 10:50:00 GMT
    • Fred:

      < SNIP .. looking for a less ugly appraoch of

      > n = 123;

      > v = dec2bin(n) - 48;

      n = 123 ;

      v = bitget(n,ceil(log2(n):-1:1))

      hth

      Jos

      #5; Wed, 07 May 2008 10:51:00 GMT
    • Jos wrote:

      > n = 123 ;

      > v = bitget(n,ceil(log2(n):-1:1))

      Interesting idea, but that particular line doesn't work.

      You got a parenthesis wrong and the case n = 2^m, m integer

      is not treated correctly. A working line would be

      v = bitget(n,floor(log2(n))+1:-1:1)

      n = 0 wouldn't work anyhow because of the logarithm.

      Thank for the help though, I wouldn't have come up with this without

      your input.

      /Fred

      #6; Wed, 07 May 2008 10:52:00 GMT
    • Fred wrote:

      >

      > Jos wrote:

      >

      > Interesting idea, but that particular line doesn't work.

      > You got a parenthesis wrong and the case n = 2^m, m integer

      > is not treated correctly. A working line would be

      > v = bitget(n,floor(log2(n))+1:-1:1)

      > n = 0 wouldn't work anyhow because of the logarithm.

      > Thank for the help though, I wouldn't have come up with this

      > without

      > your input.

      > /Fred

      and this works for n = 0 as well:

      v = bitget(n,max(ceil(log2(n+1)),1):-1:1) ;

      Jos

      #7; Wed, 07 May 2008 10:53:00 GMT
    • Jos wrote:

      > and this works for n = 0 as well:

      > v = bitget(n,max(ceil(log2(n+1)),1):-1:1) ;

      Yes it does, well done!

      The reason for asking about binary vectors in the first place, was

      that I was looking for a neat way to produce all possible nxn

      matrices with ones and zeros. My idea was to

      enumerate them with a decimal number, convert it to a binary vector,

      add extra zeros and finally reshape. Using bitget a realised that the

      procedure could be simplified.

      n = 3;

      N = 2^(n^2); %The total number of possible nxn matrices

      for k=0:N-1

      A = bitget(k,1:n^2);

      A = reshape(A,n,n);

      disp(A);

      end

      No explicit adding of zeros and no logarithm here.

      I am quite happy with this. Of couse I am doing more than just

      printing the matrices. Now, is there a way to avoid reshape by for

      example modifying the second argument in bitget to get an nxn matrix

      directly instead of a 1xn^2 vector?

      /Fred

      #8; Wed, 07 May 2008 10:54:00 GMT