Decode it.

>>> b'a string'.decode('ascii')
'a string'

To get bytes from string, encode it.

>>> 'a string'.encode('ascii')
b'a string'

Take a look at the struct module:

In [1]: import struct

In [2]: magic = b'\xcf\xfa\xed\xfe'

In [3]: decoded = struct.unpack('<I', magic)[0]

In [4]: hex(decoded)
Out[4]: '0xfeedfacf'

Use the struct module, it will also allow you to interpret the binary data in many ways; you need to define the types in a string format documented with that library:

struct.unpack('=HHf255s', bytes)

The above example expects native byte-order, two unsigned shorts, a float and a string of 255 characters.

Your code example becomes:

for each_word in logData:
    var1, var2, var3 = struct.unpack('8s16s8s', each_word)

In case you get an error TypeError: 'str' does not support the buffer interface, that is because bytes are expected, but you passed a str, so convert it to bytes and specify the encoding (in this example, UTF-8):

for each_word in logData:
    var1, var2, var3 = struct.unpack('8s16s8s', bytes(each_word, 'utf-8'))

But maybe the case that your 8/16 byte strings are long integers? In that case use the appropriate format for struct.

EDIT: it turns out you wanted to read 8 bits (not bytes), then the next 16 bits, then the next 8 bits, so you can read it as one (unsigned?) byte, one short, and another byte. The format string you should use is '=bhb' (or '=BHB' for unsigned). Example:

import struct
with open('test.bin','rb') as f:
    var1, var2, var3 = struct.unpack('=BHB', f.read(4))
    print(var1, var2, var3)

You use the built-in int() function, and pass it the base of the input number, i.e. 2 for a binary number:

>>> int('11111111', 2)
255

Here is documentation for Python 2, and for Python 3.

You need to open your file in binary mode and read only 12 bytes from your file:

import struct

with open('abc.dat', 'rb') as fobj:
    byte_string, n1, n4 = struct.unpack('4sbI', fobj.read(12)) 

You will get a byte string. Assuming it is ASCII, you can decode like this:

my_string = byte_string.decode('ascii')

The documentation of the struct contains tables of format strings. According to one of these tables a uint64 would be L.

In python you can only create it as a binary value (as a syntactic sugar), it will be converted into an integer immediately. Try it for yourself:

>>> 0b11001010
202

The same thing will happen with octal and hexadecimal values. So you can convert your binary string to an integer, with the int() function's base argument like:

>>> int('0b11001010', 2)
202

After the conversion you can do any operations on it -- just like with an integer, since it is an integer.

Of course you can convert it back at any time to a binary string, with the builtin bin() function:

>>> bin(202)
0b11001010

Just decode it before passing it to json.loads:

b = b'{\n    "begindate": "2016-11-22", \n    "enddate": "2016-11-22", \n    "guids": ["6593062E-9030-B2BC-E63A-25FBB4723ECC", \n              "5A9F8478-6673-428A-8E90-3AC4CD764543", \n              "D8243BA1-0847-48BE-9619-336CB3B3C70C"]\n}'
r = json.loads(b.decode())
print(r)
{'begindate': '2016-11-22',
 'enddate': '2016-11-22',
 'guids': ['6593062E-9030-B2BC-E63A-25FBB4723ECC',
  '5A9F8478-6673-428A-8E90-3AC4CD764543',
  'D8243BA1-0847-48BE-9619-336CB3B3C70C']}

Python 3.x makes a clear distinction between the types:

• str = '...' literals = a sequence of Unicode characters (UTF-16 or UTF-32, depending on how Python was compiled)

• bytes = b'...' literals = a sequence of octets (integers between 0 and 255)

Link for more info