undefined | Better HN

0 pointsschoen1y ago0 comments

Heh, that gives this nice expression

  sum(n>=10**i for i in range(100))

for (for example) inputs up to a googol. Or with your fix for the base case,

  1 + sum(n>=10**i for i in range(1, 100))

Another cute way that hides the loop

  from itertools import count, takewhile
  1 + len(list(takewhile(lambda x: 10**x <= n, count(1))))

0 comments

kristopolous1y ago

a way that follows the rules and also does what you're trying to do:

    def noloops(n):

      x = lambda n, digits: (n / 1e9, digits + (n >= 10) + (n >= 100) + (n >= 1e3) + (n >= 1e4) + (n >= 1e5) + (n >= 1e6) + (n >= 1e7) + (n >= 1e8) + (n >= 1e9))

      y = lambda n, digits: x(*x(*x(*x(*x(*x(n, digits))))))

      return y(*y(*y(*y(*y(*y(n, 1))))))[1]

Now we have it up to 324 digits without any loops.

psychoslave1y ago

How is that hiding the loop? I would expect takewhile to be a loop although I never used this Python facility.

schoenOP1y ago

All of the itertools functions are implemented using loops, but they heavily abstract over them so that users can think in terms of "streams" (or officially "iterators") without writing loop-oriented code themselves.

j / k navigate · click thread line to collapse