(original) (raw)

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@umbrellacode.com

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

On Tue, Jan 29, 2013 at 7:44 PM, Wolfgang Maier <wolfgang.maier@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 print\_function

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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