On 01/28/2013 01:22 PM, Wolfgang Maier wrote:

However, currently you could solve tasks like this with itertools.takewhile in the following (almost perl-like) way (I illustrate things with numbers to keep it simpler):

l=[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23] # now retrieve all numbers from 10 to 19 (combining takewhile and slicing) [n for n in itertools.takewhile(lambda n:n<20,l[len([x for x in itertools.takewhile(lambda x:x<10,l)]):])]

Nice, isn't it?

If I am not mistaken, then with my suggestion this would at least simplify to:

[n for n in l[len([x for x in l while x<10]):] while n<20]

Not great either, I admit, but at least it's fun to play this mindgame.

Well, as long as we're dreaming, how about

[n for n in l while 10 <= n < 20]

and somebody (else!) can code to skip until the first condition is met, then keep until the second condition is met, and then stop.

:)

~Ethan~