[Python-ideas] while conditional in list comprehension ?? (original) (raw)

Shane Green shane at umbrellacode.com
Wed Jan 30 02:27:52 CET 2013

Wait, it was much simpler than that…

def until(items): ... stops = [] ... def stop(): ... stops.append(1) ... yield stop ... items = iter(items) ... counter = 0 ... while not stops: ... yield next(items) ... print(counter) ... counter += 1 ...

gen = until(range(15)) stop = next(gen) [x for x in gen if x < 3 or stop()] 0 1 2 3 [0, 1, 2]

I must have just been up for too long that this looks like something new to me.

Shane Green www.umbrellacode.com 408-692-4666 | shane at umbrellacode.com

On Jan 29, 2013, at 3:37 PM, Yuriy Taraday <yorik.sar at gmail.com> wrote:

On Tue, Jan 29, 2013 at 7:44 PM, Wolfgang Maier <wolfgang.maier at biologie.uni-freiburg.de> wrote: list(i for i in range(100) if i<50 or stop()) Really (!) nice (and 2x as fast as using itertools.takewhile())!

I couldn't believe it so I had to check it: from future import printfunction import functools, itertools, operator, timeit def var1(): def gen(): for i in range(100): if i > 50: break yield i return list(gen()) def var2(): def stop(): raise StopIteration return list(i for i in range(100) if i <= 50 or stop()) def var3(): return [i for i in itertools.takewhile(lambda n: n <= 50, range(100))] def var4(): return [i for i in itertools.takewhile(functools.partial(operator.lt, 50), range(100))] if name == 'main': for f in (var1, var2, var3, var4): print(f.name, end=' ') print(timeit.timeit(f)) Results on my machine: var1 20.4974410534 var2 23.6218020916 var3 32.1543409824 var4 4.90913701057 var1 might have became the fastest of the first 3 because it's a special and very simple case. Why should explicit loops be slower that generator expressions? var3 is the slowest. I guess, because it has lambda in it. But switching to Python and back can not be faster than the last option - sitting in the C code as much as we can. -- Kind regards, Yuriy.

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