Easy function, please help.
jason.swails at gmail.com
Wed Feb 9 22:42:13 CET 2011
You've gotten several good explanations, mainly saying that 0 -> False and
not 0 -> True, which is why the while loop exits. You've also gotten advice
about how to make your method more robust (i.e. force integer division).
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). 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 (when instead it'll dump
you into an infinite loop).
You've also gotten the suggestion of typecasting to a string and then
looking at the number of characters in the string. This works fine for
integers and positive numbers, but not so well for negatives and floats,
since both the decimal and negative sign will be counted. You could
typecast to a string then strip out '-' and '.' and then count the
Or typecast to an int if you want to neglect decimals before converting to a
Or use recursion!
>>> def num_digits(n):
... if n == 0:
... return 0
... return num_digits(n//10) + 1
Why the HELL are you retuning integers in an obviously boolean
> function? Are you trying to win Pearl hacker of the year award? Please
> folks, always remember... NEVER emulate booleans, always use True/
> False for boolean behaviors.
> def is_foo():
> if something:
> return True
> return False
>>> True = 0
>>> True == False
(I know this is fixed in py3)
I think 0 -> False, 1-> True is perfectly reasonable and is fairly common in
many languages (only Fortran in the languages I use doesn't use this
Not a serious issue in any case.
Jason M. Swails
Quantum Theory Project,
University of Florida
Ph.D. Graduate Student
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