title: Is there really an @ operator in Python to calculate dot product?
body: <p>Is this answer correct: <a href="https://stackoverflow.com/a/39662710/1175080">https://stackoverflow.com/a/39662710/1175080</a> ?</p>
  
  <p>Quoting that answer.</p>
  
  <blockquote>
    <p>In Python 3.5, there is a new operator for the dot product,  so you
    can write a= A @ B  instead of a= numpy.dot(A,B)</p>
  </blockquote>
  
  <p>It does not seem to work for me.</p>
  
  <pre><code>$ python3
  Python 3.6.1 (default, Apr  4 2017, 09:40:21) 
  [GCC 4.2.1 Compatible Apple LLVM 8.1.0 (clang-802.0.38)] on darwin
  Type "help", "copyright", "credits" or "license" for more information.
  &gt;&gt;&gt; a = [1, 2, 3]
  &gt;&gt;&gt; b = [4, 5, 6]
  &gt;&gt;&gt; a @ b
  Traceback (most recent call last):
    File "&lt;stdin&gt;", line 1, in &lt;module&gt;
  TypeError: unsupported operand type(s) for @: 'list' and 'list'
  &gt;&gt;&gt;
  </code></pre>
  
  <p>But the linked answer has received 6 upvotes, so I must be missing something. Can you provide a complete example that shows how to use the <code>@</code> operator to calculate a dot product?</p>