Python - How to convert a string from binary to integer list?

I recommend the struct module.

>>> import struct
>>> s = '\x01\x02\x01\x11'
>>> struct.unpack('=4B', s)
(1, 2, 1, 17)

You can use the python int() function, the first argument is the string, the second is the base of the number.

You need to check the base, because what you posted looks like hexadecimal numbers (0x0 usually denotes a HEX number, additionally 02 is not a valid binary number).

For binary (base two):

num = int("0x11", 2); # num will be 3

For hexadecimal (base 16):

num = int("0x0A", 16); # num will be 10

To convert your string ("\0x01\0x02\0x01\0x11"):

numbers = [int(s, base) for s in "\\0x01\\0x02\\0x01\\0x11".split("\\") if len(s)]

If run with base = 16, outputs: numbers = [1, 2, 1, 17]

You can then find the average using:

average = sum(numbers)/len(numbers)

The ord() will, when given a string of length one, give you the code point of the character in that string. You should just be able to use this instead of int() in your code, which would look something like this:

sum = 0
for d in s:
    sum += ord(d)

You can do it like this:

s = "\0x01\0x02\0x01\0x11"
list = s.split('\0x')
list.remove('')

sum = 0
for d in list:
    sum += int(d)

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