Question

I have a Hex Array which looks like :

31 31 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

When converting this to binary this looks like : 0011000100110001

Each bit is a flag that relates to a number within the array. In this case this binary number would equal 2,3,7,10,11,15.

I’m not sure if there is a name for this notation, but is there any easy method to convert the hex about to get a list of decimal numbers as shown above.

So,

Each 0x31 equates to a byte or 8 bits.

Each 0x31 converts to 00110001.

The way that this binary is then supposed to be interpretted is.

0 1 2 3 4 5 6 7 8 9 10
0 0 1 1 0 0 0 1 ......

Here you can see I get the decimal values 2,3,7 from the 0x31.

Hope this makes sense. Any help would be greatly appreciated.

Answer 1

So we have the hex numbers in a space-separated string.

s = '31 31 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00'

Now we split the string, convert each byte from hex string to int (int('31', 16) == 49), then convert it to binary string (bin(49) == '0b110001'), then take away the '0b' with [2:], add zeroes at the beginning so the sequence exactly 8 long ('110001'.zfill(8) == '00110001'). Then we join all the bit strings together in one string.

s = ''.join(bin(int(b, 16))[2:].zfill(8) for b in s.split())
# Now `s` is '0011000100110001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000'

• An alternative to the above line, that works in Python 2.5:

  trans = {'0':'0000','1':'0001','2':'0010','3':'0011','4':'0100','5':'0101','6':'0110','7':'0111','8':'1000','9':'1001','a':'1010','b':'1011','c':'1100','d':'1101','e':'1110','f':'1111',' ':''}
  s = ''.join(trans[c] for c in s.lower())

Then we enumerate the bits, so every bit (b) will have a corresponding position (i), just as you described. We use a list comprehension and include only those positions at which the symbol is '1'.

r = [i for i, b in enumerate(s) if b=='1']
# Now `r` is [2, 3, 7, 10, 11, 15]