Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]

Groups > comp.lang.python > #55726 > unrolled thread

Re: Easy function, please help.

Back to article view | Back to comp.lang.python

This discussion starts older than the indexed window; earlier articles aren't shown. The article labeled Started by below is the oldest one visible, not the original post.

Contents

  Re: Easy function, please help. Michael Hrivnak <mhrivnak@hrivnak.org> - 2011-02-09 01:29 -0500
    Re: Easy function, please help. Steven D'Aprano <steve+comp.lang.python@pearwood.info> - 2011-02-10 07:08 +0000
    Re: Easy function, please help. drygal <drygalski@googlemail.com> - 2011-02-09 22:53 -0800

#55726 — Re: Easy function, please help.

Your function only works if n is an integer.  Example:

>>> num_digits(234)
3
>>> num_digits(23.4)
325

When doing integer division, python will throw away the remainder and
return an int.  Using your example of n==44, 44/10 == 4 and 4/10 == 0

Before each iteration of the while loop, the given expression (in this
case just n) is evaluated as a boolean.  Your function would act the
same if it looked like this:

def num_digits(n):
  count = 0
  while bool(n):
      count = count + 1
      n = n / 10
  return count

0 of course evaluates to False as a boolean, which is why the while loop stops.

Just for kicks, this function would work about as well:

def num_digits(n):
   return len(str(n))

And if either of these were a real function you planned to use, you'd
probably want to either cast n as an int ( int(n) ) or at least check
its type:

if not isinstance(n, int):
   raise TypeError("WTF you didn't pass me an int")

Michael

On Wed, Feb 9, 2011 at 12:52 AM, Nanderson
<mandersonrandersonanderson@gmail.com> wrote:
> def num_digits(n):
>    count = 0
>    while n:
>        count = count + 1
>        n = n / 10
>    return count
>
> This is a function that basically says how many digits are in a
> number. For example,
>>>>print num_digits(44)
> 2
>>>>print num_digits(7654)
> 4
>
> This function counts the number of decimal digits in a positive
> integer expressed in decimal format. I get this function ALMOST
> completely. The only thing I don't understand is why it eventually
> exits the loop, and goes off to the second branch. "while n" is
> confusing me. What I am thinking is that if someone puts "while n" the
> loop would be infinite. I get what is happening in the function, and I
> understand why this would work, but for some reason it's confusing me
> as to how it is exiting the loop after a certain number of times. Help
> is appreciated, thanks.
> --
> http://mail.python.org/mailman/listinfo/python-list
>

[toc] | [next] | [standalone]

#55773

On Wed, 09 Feb 2011 22:53:27 -0800, drygal wrote:

> I guess it needs:
> 
> def num_digits(n):
>    return len(str(n)) -1


I don't think so.

>>> num_digits(9)
0



-- 
Steven

[toc] | [prev] | [next] | [standalone]

#56028

On 9 Lut, 06:29, Michael Hrivnak <mhriv...@hrivnak.org> wrote:
> Your function only works if n is an integer.  Example:
>
> >>> num_digits(234)
> 3
> >>> num_digits(23.4)
>
> 325
>
> When doing integer division, python will throw away the remainder and
> return an int.  Using your example of n==44, 44/10 == 4 and 4/10 == 0
>
> Before each iteration of the while loop, the given expression (in this
> case just n) is evaluated as a boolean.  Your function would act the
> same if it looked like this:
>
> def num_digits(n):
>   count = 0
>   while bool(n):
>       count = count + 1
>       n = n / 10
>   return count
>
> 0 of course evaluates to False as a boolean, which is why the while loop stops.
>
> Just for kicks, this function would work about as well:
>
> def num_digits(n):
>    return len(str(n))
>
> And if either of these were a real function you planned to use, you'd
> probably want to either cast n as an int ( int(n) ) or at least check
> its type:
>
> if not isinstance(n, int):
>    raise TypeError("WTF you didn't pass me an int")
>
> Michael
>
> On Wed, Feb 9, 2011 at 12:52 AM, Nanderson
>
> <mandersonrandersonander...@gmail.com> wrote:
> > def num_digits(n):
> >    count = 0
> >    while n:
> >        count = count + 1
> >        n = n / 10
> >    return count
>
> > This is a function that basically says how many digits are in a
> > number. For example,
> >>>>print num_digits(44)
> > 2
> >>>>print num_digits(7654)
> > 4
>
> > This function counts the number of decimal digits in a positive
> > integer expressed in decimal format. I get this function ALMOST
> > completely. The only thing I don't understand is why it eventually
> > exits the loop, and goes off to the second branch. "while n" is
> > confusing me. What I am thinking is that if someone puts "while n" the
> > loop would be infinite. I get what is happening in the function, and I
> > understand why this would work, but for some reason it's confusing me
> > as to how it is exiting the loop after a certain number of times. Help
> > is appreciated, thanks.
> > --
> >http://mail.python.org/mailman/listinfo/python-list
>
>
I guess it needs:

def num_digits(n):
   return len(str(n)) -1

or

>>> x = lambda a: len(str(a).replace(".",""))
>>> x(23.5454)
6

Regards,
Damian.

[toc] | [prev] | [standalone]

Back to top | Article view | comp.lang.python

csiph-web