python - Convert base-2 binary number string to int

Question

I'd simply like to convert a base-2 binary number string into an int, something like this:

>>> '11111111'.fromBinaryToInt()
255

Is there a way to do this in Python?

Answer 1

You use the built-in int() function, and pass it the base of the input number, i.e. 2 for a binary number:

>>> int('11111111', 2)
255

Here is documentation for Python 2, and for Python 3.

Answer 2

Just type 0b11111111 in python interactive interface:

>>> 0b11111111
    255

Answer 3

Another way to do this is by using the bitstring module:

>>> from bitstring import BitArray
>>> b = BitArray(bin='11111111')
>>> b.uint
255

Note that the unsigned integer is different from the signed integer:

>>> b.int
-1

The bitstring module isn't a requirement, but it has lots of performant methods for turning input into and from bits into other forms, as well as manipulating them.

Answer 4

Using int with base is the right way to go. I used to do this before I found int takes base also. It is basically a reduce applied on a list comprehension of the primitive way of converting binary to decimal ( e.g. 110 = 2**0 * 0 + 2 ** 1 * 1 + 2 ** 2 * 1)

add = lambda x,y : x + y
reduce(add, [int(x) * 2 ** y for x, y in zip(list(binstr), range(len(binstr) - 1, -1, -1))])

Answer 5

If you wanna know what is happening behind the scene, then here you go.

class Binary():
    def __init__(self, binNumber):
        self._binNumber = binNumber
        self._binNumber = self._binNumber[::-1]
        self._binNumber = list(self._binNumber)
        self._x = [1]
        self._count = 1
        self._change = 2
        self._amount = 0
        print(self._ToNumber(self._binNumber))
    def _ToNumber(self, number):
        self._number = number
        for i in range (1, len (self._number)):
            self._total = self._count * self._change
            self._count = self._total
            self._x.append(self._count)
        self._deep = zip(self._number, self._x)
        for self._k, self._v in self._deep:
            if self._k == '1':
                self._amount += self._v
        return self._amount

mo = Binary('101111110')

Answer 6

A recursive Python implementation:

def int2bin(n):
    return int2bin(n >> 1) + [n & 1] if n > 1 else [1] 

Answer 7

For large matrix (10**5 rows and up) it is better to use a vectorized matmult. Pass in all rows and cols in one shot. It is extremely fast. There is no looping in python here. I originally designed it for converting many binary columns like 0/1 for like 10 different genre columns in MovieLens into a single integer for each example row.

def BitsToIntAFast(bits):
  m,n = bits.shape
  a = 2**np.arange(n)[::-1]  # -1 reverses array of powers of 2 of same length as bits
  return bits @ a

Answer 8

If you are using python3.6 or later you can use f-string to do the conversion:

Binary to decimal:

>>> print(f'{0b1011010:#0}')
90

>>> bin_2_decimal = int(f'{0b1011010:#0}')
>>> bin_2_decimal
90

binary to octal hexa and etc.

>>> f'{0b1011010:#o}'
'0o132'  # octal

>>> f'{0b1011010:#x}'
'0x5a'   # hexadecimal

>>> f'{0b1011010:#0}'
'90'     # decimal

Pay attention to 2 piece of information separated by colon.

In this way, you can convert between {binary, octal, hexadecimal, decimal} to {binary, octal, hexadecimal, decimal} by changing right side of colon[:]

:#b -> converts to binary
:#o -> converts to octal
:#x -> converts to hexadecimal 
:#0 -> converts to decimal as above example

Try changing left side of colon to have octal/hexadecimal/decimal.

Answer 9

For the record to go back and forth in basic python3:

a = 10
bin(a)
# '0b1010'

int(bin(a), 2)
# 10
eval(bin(a))
# 10