# 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?

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.

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 (`uint`) is different from the signed integer (`int`):

``````>>> b.int
-1
``````

Your question is really asking for the unsigned integer representation; this is an important distinction.

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.

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))])
``````

Just type 0b11111111 in python interactive interface:

``````>>> 0b11111111
255
``````

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')
``````

A recursive Python implementation:

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

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}'

>>> 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
:#0 -> converts to decimal as above example
``````

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

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
``````

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
``````

Here’s another concise way to do it not mentioned in any of the above answers:

``````>>> eval('0b' + '11111111')
255
``````

Admittedly, it’s probably not very fast, and it’s a very very bad idea if the string is coming from something you don’t have control over that could be malicious (such as user input), but for completeness’ sake, it does work.

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