On Mon, Mar 4, 2013 at 7:34 AM, Bryan Devaney <bryan.deva...@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. -- http://mail.python.org/mailman/listinfo/python-list
