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Re: Question on for loop

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On Mon, Mar 4, 2013 at 7:34 AM, Bryan Devaney <bryan.devaney@gmail.com> wrote:
>>         if character not in lettersGuessed:
>>
>>         return True
>>
>> return False
>
> assuming a function is being used to pass each letter of the letters guessed inside a loop itself that only continues checking if true is returned, then that could work.
>
> It is however more work than is needed.
>
> If you made secretword a list,you could just
>
> set(secretword)&set(lettersguessed)
>
> and check the result is equal to secretword.

Check the result is equal to set(secretword), I think you mean.

set(secretword).issubset(set(lettersguessed))

might be slightly more efficient, since it would not need to build and
return an intersection set.

One might also just do:

all(letter in lettersguessed for letter in secretword)

Which will be efficient if lettersguessed is already a set.

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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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