title: conda update sklearn - does not update
body: <p>I am using anaconda python</p>
  
  <pre><code>conda --v 
  conda 4.3.25
  </code></pre>
  
  <p>I am trying to update sklearn package. I followed instructions from here:
  <a href="https://stackoverflow.com/questions/36840496/conda-update-scikit-learn-also-scipy-and-numpy">conda update scikit-learn (also scipy and numpy)</a>. When I try to install the packages, it shows that it is installing sklearn 0.19</p>
  
  <pre><code>conda install scikit-learn
  The following NEW packages will be INSTALLED:
  numpy:        1.13.1-py27_0     
  scikit-learn: 0.19.0-np113py27_0
  scipy:        0.19.1-np113py27_0
  Proceed ([y]/n)? y
  </code></pre>
  
  <p>However, when I check package version, it still shows 0.15</p>
  
  <pre><code>import sklearn
  sklearn.__version__
  '0.15.2'
  </code></pre>
  
  <p>I further created environment in python.</p>
  
  <pre><code>$conda create -n sklearn python=2.7 anaconda
  </code></pre>
  
  <p>it shows below:</p>
  
  <pre><code>The following NEW packages will be INSTALLED:
  scikit-image:       0.13.0-np112py27_0
  scikit-learn:       0.18.1-np112py27_1
  
  source activate sklearn
  
  $python
  Python 2.7.13 |Anaconda 4.4.0 (x86_64)| (default, Dec 20 2016, 23:05:08) 
  [GCC 4.2.1 Compatible Apple LLVM 6.0 (clang-600.0.57)] on darwin
  Type "help", "copyright", "credits" or "license" for more information.
  Anaconda is brought to you by Continuum Analytics.
  Please check out: http://continuum.io/thanks and https://anaconda.org
  &gt;&gt;&gt; import sklearn
  &gt;&gt;&gt; sklearn.__version__
  '0.15.2'
  </code></pre>
  
  <p>Further, when I looked at packages in pycharm I see both 15.2 and 19.1. Please see attached screenshot. <a href="https://i.stack.imgur.com/zlOGz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zlOGz.png" alt="enter image description here"></a></p>
  
  <p>How to resolve this?
  I see that packages get installed in to folder ./local/lib/python2.7/site-packages. Is this expected with Conda?</p>