title: Equivalent of sorted function with compare function in Python 3
body: <p>This works perfectly in Python 2.7</p>
  
  <pre><code>def some_compare_func(x, y):
    ....
  
  a = sorted(some_list, lambda x, y: some_compare_func(x, y))
  </code></pre>
  
  <p>However, the same gives this error in Python 3.x. </p>
  
  <pre><code>TypeError: sorted expected 1 arguments, got 2
  </code></pre>
  
  <p>Is there a way to sort with a sorting function that works both in Python 2.7 and 3.x?</p>