Convert base-2 binary number string to int

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Question :

Convert base-2 binary number string to int

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 python2, and for python3.

Answered By: unwind

Answer #2:

Just type 0b11111111 in python interactive interface:

>>> 0b11111111
    255
Answered By: lengxuehx

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.

Answered By: Alex Reynolds

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))])
Answered By: Saurabh Hirani

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')
Answered By: Mohammad Mahjoub

Answer #6:

A recursive Python implementation:

def int2bin(n):
    return int2bin(n >> 1) + [n & 1] if n > 1 else [1] 
Answered By: Ludovic Trottier

Answer #7:

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.

Answered By: Robert Ranjan

Answer #8:

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
Answered By: Geoffrey Anderson

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