On Wed, Mar 25, 2015 at 2:13 PM, <otaksoftspamt...@gmail.com> wrote:
> I have a list containing 9600 integer elements - each integer is either 0 or
> 1.
>
> Starting at the front of the list, I need to combine 8 list elements into 1
> by treating them as if they were bits of one byte with 1 and 0 denoting bit
> on/off (the 8th element would be the rightmost bit of the first byte).
>
> Speed is not of utmost importance - an elegant solution is. Any suggestions?
Oooh fun!
>>> l = [1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1]
>>> list(int(''.join(str(i) for i in l),2).to_bytes(len(l)//8,'big'))
[177, 105, 117]
Convert it into a string, convert the string to an integer
(interpreting it as binary), then convert the integer into a series of
bytes, and interpret those bytes as a list of integers.
Example works in Python 3. For Python 2, you'll need ord() to get the
integers at the end.
I'm not sure how elegant this is, but it's a fun trick to play with :)
Next idea please! I love these kinds of threads.
ChrisA
--
https://mail.python.org/mailman/listinfo/python-list
