How do you express an integer as a binary number with Python literals?

I was easily able to find the answer for hex:

>>> 0x12AF
4783
>>> 0x100
256

and octal:

>>> 01267
695
>>> 0100
64

How do you use literals to express binary in Python?

Summary of Answers

• Python 2.5 and earlier: can express binary using int('01010101111',2) but not with a literal.
• Python 2.5 and earlier: there is no way to express binary literals.
• Python 2.6 beta: You can do like so: 0b1100111 or 0B1100111.
• Python 2.6 beta: will also allow 0o27 or 0O27 (second character is the letter O) to represent an octal.
• Python 3.0 beta: Same as 2.6, but will no longer allow the older 027 syntax for octals.

For reference—future Python possibilities:
Starting with Python 2.6 you can express binary literals using the prefix 0b or 0B:

>>> 0b101111
47

You can also use the new bin function to get the binary representation of a number:

>>> bin(173)
'0b10101101'

Development version of the documentation: What’s New in Python 2.6

>>> print int('01010101111',2)
687
>>> print int('11111111',2)
255

Another way.

How do you express binary literals in Python?

They’re not “binary” literals, but rather, “integer literals”. You can express integer literals with a binary format with a 0 followed by a B or b followed by a series of zeros and ones, for example:

>>> 0b0010101010
170
>>> 0B010101
21

From the Python 3 docs, these are the ways of providing integer literals in Python:

Integer literals are described by the following lexical definitions:

integer      ::=  decinteger | bininteger | octinteger | hexinteger
decinteger   ::=  nonzerodigit (["_"] digit)* | "0"+ (["_"] "0")*
bininteger   ::=  "0" ("b" | "B") (["_"] bindigit)+
octinteger   ::=  "0" ("o" | "O") (["_"] octdigit)+
hexinteger   ::=  "0" ("x" | "X") (["_"] hexdigit)+
nonzerodigit ::=  "1"..."9"
digit        ::=  "0"..."9"
bindigit     ::=  "0" | "1"
octdigit     ::=  "0"..."7"
hexdigit     ::=  digit | "a"..."f" | "A"..."F"

There is no limit for the length of integer literals apart from what can be stored in available memory.

Note that leading zeros in a non-zero decimal number are not allowed. This is for disambiguation with C-style octal literals, which Python used before version 3.0.

Some examples of integer literals:

7     2147483647                        0o177    0b100110111
3     79228162514264337593543950336     0o377    0xdeadbeef
      100_000_000_000                   0b_1110_0101

Changed in version 3.6: Underscores are now allowed for grouping purposes in literals.

Other ways of expressing binary:

You can have the zeros and ones in a string object which can be manipulated (although you should probably just do bitwise operations on the integer in most cases) – just pass int the string of zeros and ones and the base you are converting from (2):

>>> int('010101', 2)
21

You can optionally have the 0b or 0B prefix:

>>> int('0b0010101010', 2)
170

If you pass it 0 as the base, it will assume base 10 if the string doesn’t specify with a prefix:

>>> int('10101', 0)
10101
>>> int('0b10101', 0)
21

Converting from int back to human readable binary:

You can pass an integer to bin to see the string representation of a binary literal:

>>> bin(21)
'0b10101'

And you can combine bin and int to go back and forth:

>>> bin(int('010101', 2))
'0b10101'

You can use a format specification as well, if you want to have minimum width with preceding zeros:

>>> format(int('010101', 2), '{fill}{width}b'.format(width=10, fill=0))
'0000010101'
>>> format(int('010101', 2), '010b')
'0000010101'

0 in the start here specifies that the base is 8 (not 10), which is pretty easy to see:

>>> int('010101', 0)
4161

If you don’t start with a 0, then python assumes the number is base 10.

>>> int('10101', 0)
10101

As far as I can tell Python, up through 2.5, only supports hexadecimal & octal literals. I did find some discussions about adding binary to future versions but nothing definite.

I am pretty sure this is one of the things due to change in Python 3.0 with perhaps bin() to go with hex() and oct().

EDIT: lbrandy’s answer is correct in all cases.