title: TypeError: next() takes no arguments (1 given)
body: <p>I am building a SoundCloud music streaming skill for Alexa with Flask-ask in python. I have a generator function that looks like this:</p>
  
  <pre><code>def stream_likes():
     url = "http://api.soundcloud.com/me/favorites?oauth_token={}".format(oauth_token)
     r = requests.get(url)
     response = json.loads(r.text)
     for track in response:
         yield track.url
  
  
  fav_generator = stream_likes()
  </code></pre>
  
  <p>When I call <code>next(fav_generator)</code> immediately after the iterator is created, it works fine. However, when I call the function from inside of another function like this:</p>
  
  <pre><code>@ask.intent('FavoritesIntent')
  def play_favs():
      stream_url = next(fav_generator)
      return audio().play(stream_url)
  </code></pre>
  
  <p>I get the following error code.</p>
  
  <pre><code>TypeError: next() takes no arguments (1 given)
  </code></pre>
  
  <p>I would understand this error if I called a method of a class without passing the self parameter but I don't know why I am getting it in this instance.   </p>