Newbie looking for elegant solution

[Travis Griggs]

> On Mar 24, 2015, at 8:28 PM, Chris Angelico <rosuav at gmail.com> wrote:
> 
> On Wed, Mar 25, 2015 at 2:13 PM,  <otaksoftspamtrap at 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.

Me too. These are my favorite threads. Here’s my entry:

[sum(b << (7 - i) for i, b in enumerate(bits)) for bits in zip(*[l[n::8] for n in range(8)])]

I think there has to be a better way to do the left hand part, but I liked the zipped iterators on 8 slices.