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From: marduk <marduk@python.net>
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Subject: Re: Issue with my code
Date: Tue, 05 Feb 2013 17:05:16 -0500
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On Tue, Feb 5, 2013, at 04:37 PM, darnold wrote:
> On Feb 5, 2:19=C2=A0pm, maiden129 <sengokubasarafe...@gmail.com> wrote:
> > How to reverse the two loops?
> >
>=20
> s=3Dinput("Enter a string, eg(4856w23874): ")
>=20
> checkS=3D['0','1','2','3','4','5','6','7','8','9']
>=20
> for digit in checkS:
>     t =3D s.count(digit)
>     if t =3D=3D 0:
>         pass
>     elif t =3D=3D 1:
>         print(digit,"occurs 1 time.")
>     else:
>         print(digit, "occurs", t,"times.")
>=20
>=20
> >>>
> Enter a string, eg(4856w23874): 23493049weee2039412367
> 0 occurs 2 times.
> 1 occurs 1 time.
> 2 occurs 3 times.
> 3 occurs 4 times.
> 4 occurs 3 times.
> 6 occurs 1 time.
> 7 occurs 1 time.
> 9 occurs 3 times.
> >>>

Although that implementation also scans the string 10 times (s.count()),
which may not be as efficient (although it is happening in C, so perhaps
not).

A better solution involves only scanning the string once.