Convert binary string to bytearray in Python 3

Question

Despite the many related questions, I can't find any that match my problem. I'd like to change a binary string (for example, "0110100001101001") into a byte array (same example, b"hi").

I tried this:

bytes([int(i) for i in "0110100001101001"])

but I got:

b'\x00\x01\x01\x00\x01' #... and so on

What's the correct way to do this in Python 3?

Answer 1

Here's an example of doing it the first way that Patrick mentioned: convert the bitstring to an int and take 8 bits at a time. The natural way to do that generates the bytes in reverse order. To get the bytes back into the proper order I use extended slice notation on the bytearray with a step of -1: b[::-1].

def bitstring_to_bytes(s):
    v = int(s, 2)
    b = bytearray()
    while v:
        b.append(v & 0xff)
        v >>= 8
    return bytes(b[::-1])

s = "0110100001101001"
print(bitstring_to_bytes(s))

Clearly, Patrick's second way is more compact. :)

However, there's a better way to do this in Python 3: use the int.to_bytes method:

def bitstring_to_bytes(s):
    return int(s, 2).to_bytes((len(s) + 7) // 8, byteorder='big')

---

If len(s) is guaranteed to be a multiple of 8, then the first arg of .to_bytes can be simplified:

return int(s, 2).to_bytes(len(s) // 8, byteorder='big')

This will raise OverflowError if len(s) is not a multiple of 8, which may be desirable in some circumstances.

---

Another option is to use double negation to perform ceiling division. For integers a & b, floor division using //

n = a // b

gives the integer n such that
n <= a/b < n + 1
Eg,
47 // 10 gives 4, and

-47 // 10 gives -5. So

-(-47 // 10) gives 5, effectively performing ceiling division.

Thus in bitstring_to_bytes we could do:

return int(s, 2).to_bytes(-(-len(s) // 8), byteorder='big')

However, not many people are familiar with this efficient & compact idiom, so it's generally considered to be less readable than

return (s, 2).to_bytes((len(s) + 7) // 8, byteorder='big')