Bool array to integer

Just use algebra:

sum(2**i for i, v in enumerate(reversed(p)) if v)

To convert integer strings in different bases, simply use the int() function with an appropriate base.

>>> int(''.join('1' if i else '0' for i in p), 2)
233

Using generators ([1 if i else 0 for i in p]) have a more performance than map (map(int, p))here !

Using int, you can convert binary representation to int object (by specifying base 2):

>>> p = [True, True, True, False, True, False, False, True]
>>> int(''.join(map(str, map(int, p))), 2)
233

>>> p = [True, True, True, False, True, False, False, True]
>>> int("".join(map(lambda x: x and '1' or '0',p)),2)
233

int with base 2 give integer from binary

I came across this question because I wanted to do the same thing.

Mine was an X/Y problem, however: I was implementing a function and thought that I wanted to convert the input array of bits into an integer.

What I really wanted to do was to use an IntFlag from the enum built-in instead of my custom implementation.

Here's the standard example from the documentation:

>>> from enum import IntFlag
>>> class Perm(IntFlag):
...     R = 4
...     W = 2
...     X = 1
...
>>> Perm.R | Perm.W
<Perm.R|W: 6>
>>> Perm.R + Perm.W
6
>>> RW = Perm.R | Perm.W
>>> Perm.R in RW
True

It also works the other way:

>>> Perm(7)
<Perm.R|W|X: 7>

My use case was for a bitmask:

class BitMask(IntFlag):
    flag1 = 2**0
    flag2 = 2**1
    flag3 = 2**2
    flag4 = 2**3
    flag5 = 2**4
    # etcetera

With an IntFlag I can get the integer value while still retaining the useful information of what each bit represents.