Convert bytes to bits in python

问题:

I am working with Python3.2. I need to take a hex stream as an input and parse it at bit-level. So I used

bytes.fromhex(input_str)

to convert the string to actual bytes. Now how do I convert these bytes to bits?

回答1:

Another way to do this is by using the bitstring module:

>>> from bitstring import BitArray
>>> input_str = '0xff'
>>> c = BitArray(hex=input_str)
>>> c.bin
'0b11111111'

And if you need to strip the leading 0b:

>>> c.bin[2:]
'11111111'

The bitstring module isn't a requirement, as jcollado's answer shows, but it has lots of performant methods for turning input into bits and manipulating them. You might find this handy (or not), for example:

>>> c.uint
255
>>> c.invert()
>>> c.bin[2:]
'00000000'

etc.

回答2:

Operations are much faster when you work at the integer level. In particular, converting to a string as suggested here is really slow.

If you want bit 7 and 8 only, use e.g.

val = (byte >> 6) & 3

(this is: shift the byte 6 bits to the right - dropping them. Then keep only the last two bits 3 is the number with the first two bits set...)

These can easily be translated into simple CPU operations that are super fast.

回答3:

What about something like this?

>>> bin(int('ff', base=16))
'0b11111111'

This will convert the hexadecimal string you have to an integer and that integer to a string in which each byte is set to 0/1 depending on the bit-value of the integer.

As pointed out by a comment, if you need to get rid of the 0b prefix, you can do it this way:

>>> bin(int('ff', base=16)).lstrip('0b')
11111111

or this way:

>>> bin(int('ff', base=16))[2:]
11111111

回答4:

To binary:

bin(byte)[2:].zfill(8)

回答5:

I think simplest would be use numpy here. For example you can read a file as bytes and then expand it to bits easily like this:

Bytes = numpy.fromfile(filename, dtype = "uint8")
Bits = numpy.unpackbits(Bytes)

回答6:

Use ord when reading reading bytes:

byte_binary = bin(ord(f.read(1))) # Add [2:] to remove the "0b" prefix

Or

Using str.format():

'{:08b}'.format(ord(f.read(1)))

回答7:

Here how to do it using format()

print "bin_signedDate : ", ''.join(format(x, '08b') for x in bytevector)

It is important the 08b . That means it will be a maximum of 8 leading zeros be appended to complete a byte. If you don't specify this then the format will just have a variable bit length for each converted byte.

回答8:

The other answers here provide the bits in big-endian order ('\x01' becomes '00000001')

In case you're interested in little-endian order of bits, which is useful in many cases, like common representations of bignums etc - here's a snippet for that:

def bits_little_endian_from_bytes(s):
    return ''.join(bin(ord(x))[2:].rjust(8,'0')[::-1] for x in s)

And for the other direction:

def bytes_from_bits_little_endian(s):
    return ''.join(chr(int(s[i:i+8][::-1], 2)) for i in range(0, len(s), 8))

回答9:

using python format string syntax

>>> mybyte = bytes.fromhex("0F") # create my byte using a hex string
>>> binary_string = "{:08b}".format(int(mybyte.hex(),16))
>>> print(binary_string)
00001111

The second line is where the magic happens. All byte objects have a .hex() function, which returns a hex string. Using this hex string, we convert it to an integer, telling the int() function that it's a base 16 string (because hex is base 16). Then we apply formatting to that integer so it displays as a binary string. The {:08b} is where the real magic happens. It is using the Format Specification Mini-Language format_spec. Specifically it's using the width and the type parts of the format_spec syntax. The 8 sets width to 8, which is how we get the nice 0000 padding, and the b sets the type to binary.

I prefer this method over the bin() method because using a format string gives a lot more flexibility.