Chris Angelico <ros...@gmail.com> writes: > 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?
Lol. You do not. In fact, this should be syntax error :-D --- as I guess it would be if it were a lambda expression? -- https://mail.python.org/mailman/listinfo/python-list
