binary representation of an integer

[eliben]

Hello,

I'm interested in converting integers to a binary representation,
string. I.e. a desired function would produce:

dec2bin(13) => "1101"

The other way is easily done in Python with the int() function.

Perl has a very efficient way to do dec2bin, because its pack/unpack
have a B format for binary representations, so it's done with:

sub dec2bin {
my $str = unpack("B32", pack("N", shift));
$str =~ s/^0+(?=\d)//; # otherwise you'll get leading zeros
return $str;
}

Python's pack/unpack don't have the binary format for some reason, so
custom solutions have to be developed. One suggested in the ASPN
cookbook is:
http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/111286
However, it is very general and thus inefficient.

What would be the quickest way to do this ? I think that for dec2bin
conversion, using hex() and then looping with a hex->bin lookup table
would be probably much faster than the general baseconvert from the
recipe.

What do you think?

 

[Maric Michaud]

Le Tuesday 24 June 2008 10:03:58 eliben, vous avez écrit :

Hello,

I'm interested in converting integers to a binary representation,
string. I.e. a desired function would produce:

dec2bin(13) => "1101"

The other way is easily done in Python with the int() function.

Perl has a very efficient way to do dec2bin, because its pack/unpack
have a B format for binary representations, so it's done with:

sub dec2bin {
my $str = unpack("B32", pack("N", shift));
$str =~ s/^0+(?=\d)//; # otherwise you'll get leading zeros
return $str;
}

Python's pack/unpack don't have the binary format for some reason,

Yes, but I think the %b format specifier has been added to p3k but will not in
python 2.x.

so
custom solutions have to be developed. One suggested in the ASPN
cookbook is:
http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/111286
However, it is very general and thus inefficient.

What would be the quickest way to do this ? I think that for dec2bin
conversion, using hex() and then looping with a hex->bin lookup table
would be probably much faster than the general baseconvert from the
recipe.

What do you think?

Something like that, less typing with octal conversion

[8]: oct2bin =

{'0':'000', '1':'001', '2':'010', '3':'011', '4':'100', '5':'101', '6':'110', '7':'111'}

[9]: ''.join(oct2bin[e] for e in "%o"%35).lstrip('0')

...[9]: '100011'



--
_____________

Maric Michaud
_____________

Aristote - www.aristote.info
3 place des tapis
69004 Lyon
Tel: +33 4 26 88 00 97
Mobile: +33 6 32 77 00 21

 

[Mark Dickinson]

What would be the quickest way to do this ? I think that for dec2bin
conversion, using hex() and then looping with a hex->bin lookup table
would be probably much faster than the general baseconvert from the
recipe.

I suspect you're right, but it would be easy to find out: just
code up the hex->bin method and use the timeit module to do some
timings. Don't forget to strip the trailing 'L' from hex(n) if n
is a long.

If you're prepared to wait for Python 2.6, or work with the beta
version, then the conversion is already there:

Macintosh-3:trunk dickinsm$ ./python.exe
Python 2.6b1+ (trunk:64489, Jun 23 2008, 21:10:40)
[GCC 4.0.1 (Apple Inc. build 5465)] on darwin
Type "help", "copyright", "credits" or "license" for more information.'0b1101'

Interestingly, unlike hex and oct, bin doesn't add a trailing
'L' for longs:
'0b1101'

I wonder whether this is a bug...

Mark

 

[cokofreedom]

What would be the quickest way to do this ? I think that for dec2bin
conversion, using hex() and then looping with a hex->bin lookup table
would be probably much faster than the general baseconvert from the
recipe.

I suspect you're right, but it would be easy to find out: just
code up the hex->bin method and use the timeit module to do some
timings. Don't forget to strip the trailing 'L' from hex(n) if n
is a long.

If you're prepared to wait for Python 2.6, or work with the beta
version, then the conversion is already there:

Macintosh-3:trunk dickinsm$ ./python.exe
Python 2.6b1+ (trunk:64489, Jun 23 2008, 21:10:40)
[GCC 4.0.1 (Apple Inc. build 5465)] on darwin
Type "help", "copyright", "credits" or "license" for more information.>>> bin(13)

'0b1101'

Interestingly, unlike hex and oct, bin doesn't add a trailing
'L' for longs:

'0b1101'

I wonder whether this is a bug...

Mark

Strange in 2.6, but I know at least in 3.0 that all integers are C
Long's now, so the L is no longer required.

 

[cokofreedom]

And:

# return as a string
def itob_string(integer, count = 8):
return "".join(str((integer >> i) & 1) for i in range(count - 1,
-1, -1))

# return as an iterator (i.e [0, 0, 0, 0, 1, 0, 1, 0])
def itob_list(integer, count = 8):
return [(integer >> i) & 1 for i in range(count - 1, -1, -1)]

# return as a generator
def itob_generator(integer, count = 8):
return ((integer >> i) & 1 for i in range(count - 1, -1, -1))

 

[bearophileHUGS]

eliben:

Python's pack/unpack don't have the binary format for some reason, so
custom solutions have to be developed. One suggested in the ASPN
cookbook is:http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/111286
However, it is very general and thus inefficient.

Try mine, it may be fast enough for your purposes:
http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/440528

Bye,
bearophile

 

[Terry Reedy]

Strange in 2.6, but I know at least in 3.0 that all integers are C
Long's now, so the L is no longer required.

In current 2.x, the trailing L is no longer required for long integer
input; the lexer decides whether to make an int or long. Similarly,
ints are automatically converted to longs as needed instead of raising
overflow errors (as once happended). The trailing L on output would
have been removed already except for backward compatibility. But there
was no back-compatibility requirement for 0bxxxx strings.

In 3.0, all integers are class 'int'. The internal representation as
fixed or extended precision is entirely an internal implementation matter.