title: what can't I group a DataFrame by every column
body: <p>I used to perform this type of operation to eliminate duplicate records in my DataFrame</p>
  
  <pre><code>import pandas as pd
  
  data = [
      [669, 1, 2003, 2],
      [673, 1, 2003, 2],
      [675, 1, 2004, 1],
      [739, 1, 2004, 1],
      [741, 1, 2003, 1],
      [773, 1, 2004, 1],
      [791, 1, 2004, 1],
      [794, 1, 2004, 1],
  ]
  
  df = pd.DataFrame(data, columns=['A', 'B', 'C', 'D'])
  dfg = df.groupby(['A', 'B', 'C', 'D'])
  dff = dfg.first().reset_index()
  
  print pd.__version__
  print
  
  print df
  print
  print dff
  print
  </code></pre>
  
  <p>The output looks like:</p>
  
  <pre><code>0.15.1
  
       A  B     C  D
  0  669  1  2003  2
  1  673  1  2003  2
  2  675  1  2004  1
  3  739  1  2004  1
  4  741  1  2003  1
  5  773  1  2004  1
  6  791  1  2004  1
  7  794  1  2004  1
  
  Empty DataFrame
  Columns: [A, B, C, D]
  Index: []
  </code></pre>
  
  <p>You can see the dff DataFrame is empty.  I don't understand why.</p>
  
  <p>However, If I add a dummy column 'E' and group by everything else and then slice only the columns I want, all is well.</p>
  
  <pre><code>df_ = df.copy()
  df_['E'] = 'E'
  dfg_ = df_.groupby(['A', 'B', 'C', 'D'])
  dff_ = dfg_.first().reset_index()
  
  print df_[['A', 'B', 'C', 'D']]
  print
  print dff_[['A', 'B', 'C', 'D']]
  </code></pre>
  
  <p>And I get this output:</p>
  
  <pre><code>     A  B     C  D
  0  669  1  2003  2
  1  673  1  2003  2
  2  675  1  2004  1
  3  739  1  2004  1
  4  741  1  2003  1
  5  773  1  2004  1
  6  791  1  2004  1
  7  794  1  2004  1
  
       A  B     C  D
  0  669  1  2003  2
  1  673  1  2003  2
  2  675  1  2004  1
  3  739  1  2004  1
  4  741  1  2003  1
  5  773  1  2004  1
  6  791  1  2004  1
  7  794  1  2004  1
  </code></pre>
  
  <p>So my questions are these:</p>
  
  <ol>
  <li>Was this an intentional change in pandas 0.15.1?</li>
  <li>Is there a better or more appropriate way to eliminate duplicate records in my DataFrame?</li>
  </ol>
  
  <p>Thanks,
  PiR</p>