415

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.
0

8 Answers 8

370

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

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Comments

88
>>> print int('01010101111',2)
687
>>> print int('11111111',2)
255

Another way.

1 Comment

This is interesting for when you have strings, but if working with pure numbers, you add unnecessary computations to the code.
41

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'

1 Comment

int(hex(1234), 16) = 1234 :D
3

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

Comments

3

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

Comments

1

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

1 Comment

Nope, not a good method at all. You should never use eval().
-1

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.

Comments

-3

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.

Comments

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