title: 'PySide.QtCore.Signal' object has no attribute 'connect'
body: <p>I am using Python 3.4 with Pyside 1.2.4 and PyQt 4.8.7 and when I try to connect a Signal to a Slot it says:</p>
  
  <blockquote>
    <p>'PySide.QtCore.Signal' object has no attribute 'connect'</p>
  </blockquote>
  
  <p>Im am using MVC:</p>
  
  <p>Model: </p>
  
  <pre><code>from PySide.QtCore import Signal
  class Model(object):
  
      def __init__(self):
          self.updateProgress = Signal(int)
  </code></pre>
  
  <p>Controller:</p>
  
  <pre><code>class Controller(QWidget):
  """
  MVC Pattern: Represents the controller class
  
  """
  def __init__(self, parent=None):
      super().__init__(parent)
      self.model = Model()
      self.model.updateProgress.connect(self.setProgress)
  </code></pre>
  
  <p>When I look up the Class in Pycharm, by holding CTRL and click on the Signal class it looks like the following:</p>
  
  <pre><code>class Signal(object):
  """ Signal """
  def __call__(self, *args, **kwargs): # real signature unknown
      """ Call self as a function. """
      pass
  
  def __getitem__(self, *args, **kwargs): # real signature unknown
      """ Return self[key]. """
      pass
  
  def __init__(self, *args, **kwargs): # real signature unknown
      pass
  
  @staticmethod # known case of __new__
  def __new__(*args, **kwargs): # real signature unknown
      """ Create and return a new object.  See help(type) for accurate signature. """
      pass
  
  def __str__(self, *args, **kwargs): # real signature unknown
      """ Return str(self). """
      pass
  </code></pre>
  
  <p>... while there are actually supposed to be the methods <em>connect</em>, <em>disconnect</em> and <em>emit</em> according to the PySide documentation, available at:</p>
  
  <blockquote>
    <p><a href="https://srinikom.github.io/pyside-docs/PySide/QtCore/Signal.html#PySide.QtCore.Signal.connect" rel="noreferrer">https://srinikom.github.io/pyside-docs/PySide/QtCore/Signal.html#PySide.QtCore.Signal.connect</a></p>
  </blockquote>
  
  <p>Thanks for your help in advance</p>