On Monday, March 4, 2013 4:37:11 PM UTC, Ian wrote: > 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.
You are correct. sorry for the misleading answer, was digging through old shell scripts all day yesterday and brain was obviously not not the better for it. -- http://mail.python.org/mailman/listinfo/python-list
