Easy function, please help.
ethan at stoneleaf.us
Thu Feb 10 02:16:06 CET 2011
Jason Swails wrote:
> However, as surprising as this may be I'm actually with RR on this one
> (for a little) -- for code readability's sake, you should make your
> conditional more readable (i.e. don't depend on the fact that the
> iterations will take your test value down to 0 which conveniently in
> this case evaluates to False).
while n: is plenty readable. n is either something or nothing, and
something evaluates to True, nothing to False.
> This could encourage you in later cases
> to think that if this result eventually converged to a different number,
> say the multiplicative identity instead, that the same approach will
> work [...]
See above comment -- something or nothing, not mathematical identities.
> def num_digits(n):
> return len(str(n).replace('-','').replace('.',''))
> Or typecast to an int if you want to neglect decimals before converting
> to a string, etc.
> Or use recursion!
> >>> def num_digits(n):
> ... if n == 0:
> ... return 0
> ... else:
> ... return num_digits(n//10) + 1
> >>> num_digits(1)
> >>> num_digits(0)
0 is still one digit. ;)
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