On Mon, Aug 30, 2021 at 11:13 PM David Raymond <[email protected]> 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?
ChrisA
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