On Tuesday, February 5, 2013 1:56:55 PM UTC-5, marduk wrote:
> On Tue, Feb 5, 2013, at 01:38 PM, maiden129 wrote:
>
> > Hi,
>
> >
>
> > I'm trying to create this program that counts the occurrences of each
>
> > digit in a string which the user have to enter.
>
> >
>
> > Here is my code:
>
> >
>
> > s=input("Enter a string, eg(4856w23874): ")
>
> > s=list(s)
>
> >
>
> > checkS=['0','1','2','3','4','5','6','7','8','9']
>
> >
>
> > for i in s:
>
> > if i in checkS:
>
> > t=s.count(i)
>
> > if t>1:
>
> > for k in range(1,t):
>
> > s=s.remove(i)
>
> > print(i, "occurs", t,"times.")
>
> >
>
> > elif t==1:
>
> > print(i,"occurs 1 time.")
>
> > else: pass
>
> >
>
> > but it keeps showing this error:
>
> >
>
> > t=s.count(i)
>
> > AttributeError: 'NoneType' object has no attribute 'count'
>
>
>
> s=s.remove(i) does not return a new list but modifies the list in
>
> place.
>
>
>
> So you probably just want
>
>
>
> >>> s.remove(i)
>
>
>
> Also, there are various inefficiencies in your code, but that is the
>
> main issue with the AttributeError.
when I removed "s.remove(i), it starts to repeat the number of occurrences too
many times like this:
2 occurs 3 times.
2 occurs 3 times.
3 occurs 3 times.
3 occurs 3 times.
2 occurs 3 times.
2 occurs 3 times.
5 occurs 1 time.
3 occurs 3 times.
3 occurs 3 times.
4 occurs 1 time.
3 occurs 3 times.
3 occurs 3 times.
1 occurs 1 time.
2 occurs 3 times.
2 occurs 3 times.
How can I stop this?
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