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.

8 Answers

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 

I've tried this in Python 3.6.9

Convert Binary to Decimal

>>> 0b101111 47 >>> int('101111',2) 47 

Convert Decimal to binary

>>> bin(47) '0b101111' 

Place a 0 as the second parameter python assumes it as decimal.

>>> int('101111',0) 101111 

Another good method to get an integer representation from binary is to use eval()

Like so:

def getInt(binNum = 0): return eval(eval('0b' + str(n))) 

I guess this is a way to do it too. I hope this is a satisfactory answer :D

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.