Easiest way to calculate number of character in string

[Bengt Richter]

On 21 Dec 2005 15:57:35 +0100, Ove Svensson <svensson_ove at hotmail.com> wrote:

>Ove Svensson <svensson_ove at hotmail.com> writes:
>
>> "P. Schmidt-Volkmar" <no_spam at tauth.de> writes:
>> 
>> > Hi there,
>> > 
>> > I have a string in which I want to calculate how often the character  ';' 
>> > occurs. If the character does not occur 42 times, the ";" should be added so 
>> > the 42 are reached.
>> > 
>> > My solution is slow and wrong:
>> >     for Position in range (0, len(Zeile)):
>> >         if Zeile[Position]==';': AnzahlSemikolon = AnzahlSemikolon +1
>> >     if AnzahlSemikolon < 42:
>> >         for Zaehler in range(AnzahlSemikolon, 42):
>> >             Zeile = Zeile + ';'
>> >         Dreckskram = Dreckskram +1
>> > 
>> > How can this be achieved easily?
>> > 
>> > Thanks,
>> > 
>> > Pascal 
>> > 
>> > 
>> 
>> What about this:
>> 
>> Zaehler += ';'*max(0,42-Zaehler.count(';'))
>> 
>
>Sorry, should have been
>
>Zeile += ';'*max(0,42-Zeile.count(';'))

I think You don't need the max

 >>> for n in xrange(-3,4): print '%3s: %r'%(n, n*';')
 ...
  -3: ''
  -2: ''
  -1: ''
   0: ''
   1: ';'
   2: ';;'
   3: ';;;'

I.e.,

    Zeile += ';'*(42-Zeile.count(';'))

should work, since a string is a sequence type and

   http://docs.python.org/lib/typesseq.html

Says
"""
    Operation       Result                                Notes
    ...
    s * n , n * s   n shallow copies of s concatenated    (2)
...
(2)
    Values of n less than 0 are treated as 0 (which yields an empty sequence of the same type as s). ...
"""

I guess it might be nice to mention this in help(str) also, to publish a useful fact better.
Maybe under str.__mul__ ?

Regards,
Bengt Richter