convert integer to binary

[Bengt Richter]

On 02 May 2003 15:42:21 -0700, Paul Rubin <http://phr.cx@NOSPAM.invalid> wrote:

>ekranawetter at utanet.at (ekranawetter-piber) writes:
>
>> Hi all,
>> 
>> I couldn't find any function in Python to convert integers to binary
>> and vice versa, for example:
>> 
>> integer 16 = bin 10000, or 
>> integer 15 = bin  1111
>> bin 11 = integer 3
>
>Note this doesn't bother stripping leading zeros from the result:
>
>def i2b(n):
>  hdigits = ('0000','0001','0010','0011','0100','0101','0110','0111',
>             '1000','1001','1010','1011','1100','1101','1110','1111')
>
>  a = [hdigits[int(d,16)] for d in hex(n)[2:]]
>  return ''.join(a)

Stripping leading zeroes is easy enough
   return ''.join(a).lstrip('0')

... digs in dustbin... (first bit is explicit 2's complement sign bit here)

 >>> def binstr(n, maxbits=None):
 ...     absn = abs(n)
 ...     s=[chr(((n>>b)&1)+48) for b in xrange(maxbits or len(hex(n))*4)
 ...         if maxbits or not b or absn>>(b-1)]
 ...     s.reverse()
 ...     return ''.join(s)
 ...
 >>> binstr(15)
 '01111'
 >>> binstr(16)
 '010000'
 >>> binstr(-15)
 '10001'
 >>> binstr(-16)
 '110000'
 >>> binstr(-16,32)
 '11111111111111111111111111110000'
 >>> for i in range(-9,10): print binstr(i),
 ...
 10111 11000 1001 1010 1011 1100 101 110 11 0 01 010 011 0100 0101 0110 0111 01000 01001

BTW, len(hex(n))*4 is just to get a sufficient count, not an exact one, so the 'L'
on the end of a long doesn't hurt except a few extra loops in the list comp that don't
add anything.

BTW2, anybody for binary literals of the form sbxxxx, where s is the sign bit as above
and xxxx is the rest? Similar to a proposed hex format of 0xnnnn and 1xnnn to indicate
what sign bit to extend above the bottom bits. Works nicely for unified ints and longs.

Regards,
Bengt Richter