python bytes to bit string

Question

I have value of the type bytes that need to be converted to BIT STRING

bytes_val = (b'\x80\x00', 14)

the bytes in index zero need to be converted to bit string of length as indicated by the second element (14 in this case) and formatted as groups of 8 bits like below.

expected output => '10000000 000000'B

Another example

bytes_val2 = (b'\xff\xff\xff\xff\xf0\x00', 45) #=> '11111111 11111111 11111111 11111111 11110000 00000'B

• Błotosmętek: ...and the 14...?
• kaya3: Does this answer your question? Whats a simple way to turn bytes into a binary string?
• Błotosmętek: Does this answer your question? Convert string to binary in python
• user3206440: @Błotosmętek - 14 is length of the expected bit string !!
• Boseong Choi: Should (b'\x80\x01', 14) produces also '1000000 000000'B?

Answer 1

What about some combination of formatting (below with f-string but can be done otherwise), and slicing:

def bytes2binstr(b, n=None):
 s = ' '.join(f'{x:08b}' for x in b)
 return s if n is None else s[:n + n // 8 + (0 if n % 8 else -1)]

If I understood correctly (I am not sure what the B at the end is supposed to mean), it passes your tests and a couple more:

func = bytes2binstr
args = (
 (b'\x80\x00', None),
 (b'\x80\x00', 14),
 (b'\x0f\x00', 14),
 (b'\xff\xff\xff\xff\xf0\x00', 16),
 (b'\xff\xff\xff\xff\xf0\x00', 22),
 (b'\x0f\xff\xff\xff\xf0\x00', 45),
 (b'\xff\xff\xff\xff\xf0\x00', 45),
)
for arg in args:
 print(arg)
 print(repr(func(*arg)))
# (b'\x80\x00', None)
# '10000000 00000000'
# (b'\x80\x00', 14)
# '10000000 000000'
# (b'\x0f\x00', 14)
# '00001111 000000'
# (b'\xff\xff\xff\xff\xf0\x00', 16)
# '11111111 11111111'
# (b'\xff\xff\xff\xff\xf0\x00', 22)
# '11111111 11111111 111111'
# (b'\x0f\xff\xff\xff\xf0\x00', 45)
# '00001111 11111111 11111111 11111111 11110000 00000'
# (b'\xff\xff\xff\xff\xf0\x00', 45)
# '11111111 11111111 11111111 11111111 11110000 00000'

---

Explanation

• we start from a bytes object
• iterating through it gives us a single byte as a number
• each byte is 8 bit, so decoding that will already give us the correct separation
• each byte is formatted using the b binary specifier, with some additional formatting: 0 zero fill, 8 minimum length
• we join (concatenate) the result of the formatting using ' ' as "separator"
• finally the result is returned as is if a maximum number of bits n was not specified (set to None), otherwise the result is cropped to n + the number of spaces that were added in-between the 8-character groups.

---

In the solution above 8 is somewhat hard-coded. If you want it to be a parameter, you may want to look into (possibly a variation of) @kederrac first answer using int.from_bytes(). This could look something like:

def bytes2binstr_frombytes(b, n=None, k=8):
 s = '{x:0{m}b}'.format(m=len(b) * 8, x=int.from_bytes(b, byteorder='big'))[:n]
 return ' '.join([s[i:i + k] for i in range(0, len(s), k)])

which gives the same output as above.

Speedwise, the int.from_bytes()-based solution is also faster:

for i in range(2, 7):
 n = 10 ** i
 print(n)
 b = b''.join([random.randint(0, 2 ** 8 - 1).to_bytes(1, 'big') for _ in range(n)])
 for func in funcs:
 print(func.__name__, funcs[0](b, n * 7) == func(b, n * 7))
 %timeit func(b, n * 7)
 print()
# 100
# bytes2binstr True
# 10000 loops, best of 3: 33.9 μs per loop
# bytes2binstr_frombytes True
# 100000 loops, best of 3: 15.1 μs per loop
# 1000
# bytes2binstr True
# 1000 loops, best of 3: 332 μs per loop
# bytes2binstr_frombytes True
# 10000 loops, best of 3: 134 μs per loop
# 10000
# bytes2binstr True
# 100 loops, best of 3: 3.29 ms per loop
# bytes2binstr_frombytes True
# 1000 loops, best of 3: 1.33 ms per loop
# 100000
# bytes2binstr True
# 10 loops, best of 3: 37.7 ms per loop
# bytes2binstr_frombytes True
# 100 loops, best of 3: 16.7 ms per loop
# 1000000
# bytes2binstr True
# 1 loop, best of 3: 400 ms per loop
# bytes2binstr_frombytes True
# 10 loops, best of 3: 190 ms per loop

• Martijn Pieters: String formatting has a perfectly serviceable binary formatting option for integers, one that doesn’t require slicing of a prefix. Why not use that?
• norok2: @MartijnPieters I was assuming that > and 0 specifier would not have worked for b, but I was wrong. Thanks for the improvement!
• Martijn Pieters: You don't need > at all, it's the default alignment for numbers (each x in the bytes object b is an integer in the interval [0, 256)).

Answer 2

you can use:

def bytest_to_bit(by, n):
 bi = "{:0{l}b}".format(int.from_bytes(by, byteorder='big'), l=len(by) * 8)[:n]
 return ' '.join([bi[i:i + 8] for i in range(0, len(bi), 8)])
bytest_to_bit(b'\xff\xff\xff\xff\xf0\x00', 45)

output:

'11111111 11111111 11111111 11111111 11110000 00000'

steps:

1. transform your bytes to an integer using int.from_bytes

2. str.format method can take a binary format spec.

---

also, you can use a more compact form where each byte is formatted:

def bytest_to_bit(by, n):
 bi = ' '.join(map('{:08b}'.format, by))
 return bi[:n + len(by) - 1].rstrip()
bytest_to_bit(b'\xff\xff\xff\xff\xf0\x00', 45)

• norok2: Eventually, you may need some extra formatting to handle corner cases for inputs whose binary representation does not start with a 1. Not specified in the question, though, so not 100% sure, but e.g. (b'0円f00円', 14) would not play out well if the required output must have leading 0s.
• Martijn Pieters: Without a minimum width leading zeros will be dropped from your output, so (b'\x00\x00', 16) produces '0', not '0000000000000000'. You would want to use "{:0{l}b}".format(int.from_bytes(by, byteorder='big'), l=len(by) * 8)[:n].

Answer 3

test_data = [
 (b'\x80\x00', 14),
 (b'\xff\xff\xff\xff\xf0\x00', 45),
]
def get_bit_string(bytes_, length) -> str:
 output_chars = []
 for byte in bytes_:
 for _ in range(8):
 if length <= 0:
 return ''.join(output_chars)
 output_chars.append(str(byte >> 7 & 1))
 byte <<= 1
 length -= 1
 output_chars.append(' ')
 return ''.join(output_chars)
for data in test_data:
 print(get_bit_string(*data))

output:

10000000 000000
11111111 11111111 11111111 11111111 11110000 00000

explanation:

• length: Start from target legnth, and decreasing to 0.
• if length <= 0: return ...: If we reached target length, stop and return.
• ''.join(output_chars): Make string from list.
• str(byte >> 7 & 1)
  • byte >> 7: Shift 7 bits to right(only remains MSB since byte has 8 bits.)
  • MSB means Most Significant Bit
  • (...) & 1: Bit-wise and operation. It extracts LSB.
• byte <<= 1: Shift 1 bit to left for byte.
• length -= 1: Decreasing length.

• user3206440: would you please add some descriptions on what the different lines do? - Thanks !!
• Boseong Choi: I appended explanation :)

Answer 4

This is lazy version.
It neither loads nor processes the entire bytes.
This one does halt regardless of input size.
The other solutions may not!

I use collections.deque to build bit string.

from collections import deque
from itertools import chain, repeat, starmap
import os 
def bit_lenght_list(n):
 eights, rem = divmod(n, 8)
 return chain(repeat(8, eights), (rem,))
def build_bitstring(byte, bit_length):
 d = deque("0" * 8, 8)
 d.extend(bin(byte)[2:])
 return "".join(d)[:bit_length]
def bytes_to_bits(byte_string, bits):
 return "{!r}B".format(
 " ".join(starmap(build_bitstring, zip(byte_string, bit_lenght_list(bits))))
 )

Test;

In [1]: bytes_ = os.urandom(int(1e9)) 
In [2]: timeit bytes_to_bits(bytes_, 0) 
4.21 μs ± 27.8 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [3]: timeit bytes_to_bits(os.urandom(1), int(1e9)) 
6.8 μs ± 51 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [4]: bytes_ = os.urandom(6) 
In [5]: bytes_ 
Out[5]: b'\xbf\xd5\x08\xbe$\x01'
In [6]: timeit bytes_to_bits(bytes_, 45) #'10111111 11010101 00001000 10111110 00100100 00000'B 
12.3 μs ± 85 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [7]: bytes_to_bits(bytes_, 14) 
Out[7]: "'10111111 110101'B"

Answer 5

when you say BIT you mean binary? I would try

bytes_val = b'\\x80\\x00'
for byte in bytes_val:
 value_in_binary = bin(byte)

• user3206440: BIT String - string representation of binary.

Answer 6

This gives the answer without python's binary representation pre-fixed 0b:

bit_str = ' '.join(bin(i).replace('0b', '') for i in bytes_val)

• user3206440: getting TypeError: 'bytes' object cannot be interpreted as an integer
• Błotosmętek: Is that Python 2.x or 3.x?
• user3206440: >>> sys.version_info sys.version_info(major=3, minor=6, micro=8, releaselevel='final', serial=0) >>> bytes_val = (b'\x80\x00', 14) >>> ' '.join(bin(i).replace('0b', '') for i in bytes_val) Traceback (most recent call last): File "<stdin>", line 1, in <module> File "<stdin>", line 1, in <genexpr> TypeError: 'bytes' object cannot be interpreted as an integer >>>

Answer 7

This works in Python 3.x:

def to_bin(l):
 val, length = l
 bit_str = ''.join(bin(i).replace('0b', '') for i in val)
 if len(bit_str) < length:
 # pad with zeros
 return '0'*(length-len(bit_str)) + bit_str
 else:
 # cut to size
 return bit_str[:length]
bytes_val = [b'\x80\x00',14]
print(to_bin(bytes_val))

and this works in 2.x:

def to_bin(l):
 val, length = l
 bit_str = ''.join(bin(ord(i)).replace('0b', '') for i in val)
 if len(bit_str) < length:
 # pad with zeros
 return '0'*(length-len(bit_str)) + bit_str
 else:
 # cut to size
 return bit_str[:length]
bytes_val = [b'\x80\x00',14]
print(to_bin(bytes_val))

Both produce result 00000100000000