On Wed, Mar 25, 2015 at 3:04 PM, Paul Rubin <no.email@nospam.invalid> wrote:
> This works for me in Python 2.7 but I think
> Python 3 gratuitously broke tuple unpacking so it won't work there:
>
> ================================================================
>
> from itertools import count, groupby
> old = [0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1]
> new = [reduce(lambda x,(y,i):x*2+y, g, 0)
> for k,g in groupby(zip(old,count()), lambda (a,b): b//8)]
> print new
>
>>>> [18, 222, 53]
> ================================================================
You don't need tuple unpacking. Here's the Py3 version of the above:
from functools import reduce
new = [reduce(lambda x,y:x*2+y[0], g, 0)
for k,g in groupby(zip(old,count()), lambda a: a[1]//8)]
ChrisA
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