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Re: Question on for loop

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In fact this code is already doing what you want, but if the second
character, by example, is not in secrectWord it'll jump out of the for and
return. If you want that interact through the all characters and  maybe
count how many them are in the secrectWord, just take of the return there
or do some list comprehension like: [ r for r in secrecWord if r in
lettersGuessed] .



[]'s



On Mon, Mar 4, 2013 at 9:18 AM, newtopython <roshen.sethna@gmail.com> wrote:

> Hi all,
>
> I'm super new to python, just fyi.
>
> In the piece of code below, secretWord is a string and lettersGuessed is a
> list. I'm trying to find out if ALL the characters of secretWord are
> included in lettersGuessed, even if there are additional values in the
> lettersGuessed list that aren't in secretWord.
>
> What this code is doing is only checking the first character of secretWord
> and then returning True or False. How do I get it to iterate through ALL of
> the characters of secretWord?
>
> for character in secretWord:
>         if character not in lettersGuessed:
>         return True
> return False
>
> Thanks!
>
> Ro
> --
> http://mail.python.org/mailman/listinfo/python-list
>



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Question on for loop newtopython <roshen.sethna@gmail.com> - 2013-03-04 04:18 -0800
  Re: Question on for loop leo kirotawa <kirotawa@gmail.com> - 2013-03-04 09:59 -0300
  Re: Question on for loop Joel Goldstick <joel.goldstick@gmail.com> - 2013-03-04 08:04 -0500
  Re: Question on for loop Dave Angel <davea@davea.name> - 2013-03-04 08:36 -0500
  Re: Question on for loop Bryan Devaney <bryan.devaney@gmail.com> - 2013-03-04 06:34 -0800
    Re: Question on for loop Ian Kelly <ian.g.kelly@gmail.com> - 2013-03-04 09:37 -0700
      Re: Question on for loop Bryan Devaney <bryan.devaney@gmail.com> - 2013-03-05 03:12 -0800
      Re: Question on for loop Bryan Devaney <bryan.devaney@gmail.com> - 2013-03-05 03:12 -0800
  Re: Question on for loop Rick Johnson <rantingrickjohnson@gmail.com> - 2013-03-04 07:41 -0800
  Re: Question on for loop Ricardo Aráoz <ricaraoz@gmail.com> - 2013-03-04 10:21 -0300

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