On Sunday, May 5, 2013 9:21:33 PM UTC-4, alex23 wrote: > On May 6, 10:59 am, Bradley Wright <bradley.wright....@gmail.com> > > wrote: > > > def fizz_cout(x): > > > count = 0 > > > for item in x: > > > while item == "fizz": > > > count += 1 > > > return count > > > > > > Please remember that i am a eager beginner, where am i going wrong? > > > > There are several problems with your code: > > > > > for item in x: > > > while item == "fizz": > > > count += 1 > > > > The `for` takes an item out of the list `x`. If that item is the > > string 'fizz', it increments count. As it's a `while` loop, it will > > continue to increment for as long as `item` is 'fizz'. Since the while > > loop doesn't look up another list item, it will remain as 'fizz' until > > the end of time. Well, it would except for your second bug: > > > > > while item == "fizz": > > > count += 1 > > > return count > > > > The very first time it encounters a list item that is 'fizz', it adds > > one to `count`, then exits the function passing back `count`. > > > > You want to move the return to _outside_ the for loop, and you want to > > change your `while` condition to an `if` instead.
Thank you Alex - much appreciated, about to implement right now! -- http://mail.python.org/mailman/listinfo/python-list
