Fast recursive generators?

[Gabriel Genellina]

En Fri, 28 Oct 2011 15:10:14 -0300, Michael McGlothlin  
<michaelm at plumbersstock.com> escribió:

> I'm trying to generate a list of values where each value is dependent
> on the previous value in the list and this bit of code needs to be
> repeatedly so I'd like it to be fast. It doesn't seem that
> comprehensions will work as each pass needs to take the result of the
> previous pass as it's argument. map() doesn't seem likely. filter() or
> reduce() seem workable but not very clean. Is there a good way to do
> this? About the best I can get is this:
>
> l = [ func ( start ) ]
> f = lambda a: func ( l[-1] ) or a
> filter ( f, range ( big_number, -1, -1 ) )
>
>
> I guess I'm looking for something more like:
>
> l = do ( lambda a: func ( a ), big_number, start )

What about a generator function?

def my_generator():
   prev = 1
   yield prev
   while True:
     this = 2*prev
     yield this
     prev = this

print list(itertools.islice(my_generator(), 10))

-- 
Gabriel Genellina