BCD codes work with 4 bits per number and usually only encode the digits 0 - 9. Thus, each byte in your sequence contains 2 numbers, 1 for 4 bits of information.
The following method uses a generator to create these numbers; I assume that a value of 0xF means that there are no more digits:
def bcdDigits(chars): for char in chars: char = ord(char) for val in (char >> 4, char & 0xF): if val == 0xF: return yield val
Here I use the shift operator to the right to move the leftmost 4 bits to the right, and the bitwise AND to select only the rightmost 4 bits.
Demonstration:
>>> characters = ('3', '\x00', '\x02', '\x05', '\x15', '\x13', 'G', 'O', '\xff', '\xff', '\xff', '\xff', '\xff', '\xff', '\xff', '\xff') >>> list(bcdDigits(characters)) [3, 3, 0, 0, 0, 2, 0, 5, 1, 5, 1, 3, 4, 7, 4]
The method works with output c ; you can skip the ord call in the method if you pass integers directly (but use unsigned option B instead). Alternatively, you can simply read these 16 bytes directly from your file and apply this function to these bytes directly without using a struct.