title: Rename duplicated index values pandas DataFrame
body: <p>I have a DataFrame that contains some duplicated index values:</p>
  
  <pre><code>df1 =  pd.DataFrame( np.random.randn(6,6), columns = pd.date_range('1/1/2010', periods=6), index = {"A", "B", "C", "D", "E", "F"})
  df1.rename(index = {"C": "A", "B": "E"}, inplace = 1)
  
  ipdb&gt; df1
        2010-01-01  2010-01-02  2010-01-03  2010-01-04  2010-01-05  2010-01-06
   A   -1.163883    0.593760    2.323342   -0.928527    0.058336   -0.209101
   A   -0.593566   -0.894161   -0.789849    1.452725    0.821477   -0.738937
   E   -0.670305   -1.788403    0.134790   -0.270894    0.672948    1.149089
   F    1.707686    0.323213    0.048503    1.168898    0.002662   -1.988825
   D    0.403028   -0.879873   -1.809991   -1.817214   -0.012758    0.283450
   E   -0.224405   -1.803301    0.582946    0.338941    0.798908    0.714560
  </code></pre>
  
  <p>I would like to change only the name of the duplicated values and to obtain a DataFrame like the following one:</p>
  
  <pre><code>ipdb&gt; df1
       2010-01-01  2010-01-02  2010-01-03  2010-01-04  2010-01-05  2010-01-06
  A   -1.163883    0.593760    2.323342   -0.928527    0.058336   -0.209101
  A_dp   -0.593566   -0.894161   -0.789849    1.452725    0.821477   -0.738937
  E   -0.670305   -1.788403    0.134790   -0.270894    0.672948    1.149089
  F    1.707686    0.323213    0.048503    1.168898    0.002662   -1.988825
  D    0.403028   -0.879873   -1.809991   -1.817214   -0.012758    0.283450
  E_dp   -0.224405   -1.803301    0.582946    0.338941    0.798908    0.714560
  </code></pre>
  
  <p><strong>My approach:</strong></p>
  
  <p>(i) Create dictionary with new names</p>
  
  <pre><code>old_names = df1[df1.index.duplicated()].index.values
  new_names = df1[df1.index.duplicated()].index.values + "_dp"
  dictionary = dict(zip(old_names, new_names))
  </code></pre>
  
  <p>(ii) Rename only the duplicated values</p>
  
  <pre><code>df1.loc[df1.index.duplicated(),:].rename(index = dictionary, inplace = True)
  </code></pre>
  
  <p>However this does not seem to work. </p>