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Subject: Re: Newbie looking for elegant solution
From: Travis Griggs <travisgriggs@gmail.com>
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Date: Wed, 25 Mar 2015 11:49:56 -0700
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> On Mar 24, 2015, at 8:28 PM, Chris Angelico <rosuav@gmail.com> wrote:
>=20
> On Wed, Mar 25, 2015 at 2:13 PM,  <otaksoftspamtrap@gmail.com> wrote:
>> I have a list containing 9600 integer elements - each integer is =
either 0 or 1.
>>=20
>> 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).
>>=20
>> Speed is not of utmost importance - an elegant solution is. Any =
suggestions?
>=20
> Oooh fun!
>=20
>>>> l =3D [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]
>=20
> 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.
>=20
> Example works in Python 3. For Python 2, you'll need ord() to get the
> integers at the end.
>=20
> I'm not sure how elegant this is, but it's a fun trick to play with :)
>=20
> Next idea please! I love these kinds of threads.

Me too. These are my favorite threads. Here=E2=80=99s 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.