title: convertion of datetime to numpy datetime without timezone info
body: <p>Suppose I have a <code>datetime</code> variable: </p>
  
  <pre><code>     dt = datetime.datetime(2001,1,1,0,0)
  </code></pre>
  
  <p>and I convert it to numpy as follows <code>numpy.datetime64(dt)</code> I get</p>
  
  <pre><code>     numpy.datetime64('2000-12-31T19:00:00.000000-0500')
  </code></pre>
  
  <p>with <code>dtype('&lt;M8[us]')</code></p>
  
  <p>But this automatically takes into account my time-zone (i.e. EST in this case) and gives me back a date of 2001-12-31 and a time of 19:00 hours. </p>
  
  <p>How can I convert it to <code>datetime64[D]</code> in numpy that ignores the timezone information and simply gives me </p>
  
  <pre><code>     numpy.datetime64('2001-01-01')
  </code></pre>
  
  <p>with <code>dtype('&lt;M8[D]')</code></p>
  
  <p><a href="http://docs.scipy.org/doc/numpy/reference/arrays.datetime.html" rel="nofollow">The numpy datetime64 doc page</a> gives no information on how to ignore the time-zone or give the default time-zone as UTC</p>