On 31/08/2021 01.50, Chris Angelico wrote:
> On Mon, Aug 30, 2021 at 11:13 PM David Raymond <david.raym...@tomtom.com>
> wrote:
>>
>>> def how_many_times():
>>> x, y = 0, 1
>>> c = 0
>>> while x != y:
>>> c = c + 1
>>> x, y = roll()
>>> return c, (x, y)
>>
>> Since I haven't seen it used in answers yet, here's another option using our
>> new walrus operator
>>
>> def how_many_times():
>> roll_count = 1
>> while (rolls := roll())[0] != rolls[1]:
>> roll_count += 1
>> return (roll_count, rolls)
>>
>
> Since we're creating solutions that use features in completely
> unnecessary ways, here's a version that uses collections.Counter:
>
> def how_many_times():
> return next((count, rolls) for count, rolls in
> enumerate(iter(roll, None)) if len(Counter(rolls)) == 1)
>
> Do I get bonus points for it being a one-liner that doesn't fit in
> eighty characters?
Herewith my claim to one-liner fame (assuming such leads in any way to
virtue or fame)
It retains @Peter's preference for a more re-usable roll_die() which
returns a single event, cf the OP's roll() which returns two results).
import itertools, random
def roll_die():
while True:
yield random.randrange(1, 7)
def how_many_times():
return list( itertools.takewhile( lambda r:r[ 0 ] != r[ 1 ],
zip( roll_die(), roll_die() )
)
)
Also, a claim for 'bonus points' because the one-liner will fit within
80-characters - if only I didn't have that pernicious and vile habit of
coding a more readable layout.
It doesn't use a two-arg iter, but still rates because it does use a
relatively-obscure member of the itertools library...
https://docs.python.org/3.8/library/itertools.html#itertools.takewhile
--
Regards,
=dn
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