Using Python How can I read the bits in a byte?

Question

I have a file where the first byte contains encoded information. In Matlab I can read the byte bit by bit with var = fread(file, 8, \'ubit1\'), and then retrie

Answer 1

There are two possible ways to return the i-th bit of a byte. The "first bit" could refer to the high-order bit or it could refer to the lower order bit.

Here is a function that takes a string and index as parameters and returns the value of the bit at that location. As written, it treats the low-order bit as the first bit. If you want the high order bit first, just uncomment the indicated line.

def bit_from_string(string, index):
       i, j = divmod(index, 8)

       # Uncomment this if you want the high-order bit first
       # j = 8 - j

       if ord(string[i]) & (1 << j):
              return 1
       else:
              return 0

The indexing starts at 0. If you want the indexing to start at 1, you can adjust index in the function before calling divmod.

Example usage:

>>> for i in range(8):
>>>       print i, bit_from_string('\x04', i)
0 0
1 0
2 1
3 0
4 0
5 0
6 0
7 0

Now, for how it works:

A string is composed of 8-bit bytes, so first we use divmod() to break the index into to parts:

• i: the index of the correct byte within the string
• j: the index of the correct bit within that byte

We use the ord() function to convert the character at string[i] into an integer type. Then, (1 << j) computes the value of the j-th bit by left-shifting 1 by j. Finally, we use bitwise-and to test if that bit is set. If so return 1, otherwise return 0.

Answer 2

The smallest unit you'll be able to work with is a byte. To work at the bit level you need to use bitwise operators.

x = 3
#Check if the 1st bit is set:
x&1 != 0
#Returns True

#Check if the 2nd bit is set:
x&2 != 0
#Returns True

#Check if the 3rd bit is set:
x&4 != 0
#Returns False

Answer 3

To read a byte from a file: bytestring = open(filename, 'rb').read(1). Note: the file is opened in the binary mode.

To get bits, convert the bytestring into an integer: byte = bytestring[0] (Python 3) or byte = ord(bytestring[0]) (Python 2) and extract the desired bit: (byte >> i) & 1:

>>> for i in range(8): (b'a'[0] >> i) & 1
... 
1
0
0
0
0
1
1
0
>>> bin(b'a'[0])
'0b1100001'

Answer 4

Joining some of the previous answers I would use:

[int(i) for i in "{0:08b}".format(byte)]

For each byte read from the file. The results for an 0x88 byte example is:

>>> [int(i) for i in "{0:08b}".format(0x88)]
[1, 0, 0, 0, 1, 0, 0, 0]

You can assign it to a variable and work as per your initial request. The "{0.08}" is to guarantee the full byte length

Answer 5

This is pretty fast I would think:

import itertools
data = range(10)
format = "{:0>8b}".format
newdata = (False if n == '0' else True for n in itertools.chain.from_iterable(map(format, data)))
print(newdata) # prints tons of True and False

Answer 6

You won't be able to read each bit one by one - you have to read it byte by byte. You can easily extract the bits out, though:

f = open("myfile", 'rb')
# read one byte
byte = f.read(1)
# convert the byte to an integer representation
byte = ord(byte)
# now convert to string of 1s and 0s
byte = bin(byte)[2:].rjust(8, '0')
# now byte contains a string with 0s and 1s
for bit in byte:
    print bit