4

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
user3206440
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7 Answers7

6

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
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norok2
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  • String formatting has a perfectly serviceable binary formatting option for integers, one that doesn’t require slicing of a prefix. Why not use that? – Martijn Pieters Apr 08 '20 at 21:33
  • @MartijnPieters I was assuming that `>` and `0` specifier would not have worked for `b`, but I was wrong. Thanks for the improvement! – norok2 Apr 09 '20 at 06:09
  • 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)`). – Martijn Pieters Apr 09 '20 at 10:51
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)
kederrac
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  • 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'\0f\00', 14)` would not play out well if the required output must have leading `0`s. – norok2 Apr 09 '20 at 09:19
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    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]`. – Martijn Pieters Apr 09 '20 at 10:56
0
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.
Boseong Choi
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0

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"
Nizam Mohamed
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-1

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)
Roni Antonio
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-1

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

bit_str = ' '.join(bin(i).replace('0b', '') for i in bytes_val)
chuck
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  • getting `TypeError: 'bytes' object cannot be interpreted as an integer` – user3206440 Mar 08 '20 at 06:10
  • Is that Python 2.x or 3.x? – Błotosmętek Mar 08 '20 at 09:13
  • `>>> 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 "", line 1, in File "", line 1, in TypeError: 'bytes' object cannot be interpreted as an integer >>>` – user3206440 Apr 04 '20 at 01:57
-1

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

Błotosmętek
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