title: Pandas Column Multiindex Subtracting Columns from each other
body: <p><strong>pandas DataFrame:</strong></p>
  
  <p>Constructor:</p>
  
  <pre><code>c = pd.MultiIndex.from_product([['AAPL','AMZN'],['price','custom']])
  i = pd.date_range(start='2017-01-01',end='2017-01-6')
  df1 = pd.DataFrame(index=i,columns=c)
  
  df1.loc[:,('AAPL','price')] = list(range(51,57))
  df1.loc[:,('AMZN','price')] = list(range(101,107))
  df1.loc[:,('AAPL','custom')] = list(range(1,7))
  df1.loc[:,('AMZN','custom')] = list(range(17,23))
  df1.index.set_names('Dates',inplace=True)
  df1.sort_index(axis=1,level=0,inplace=True) # needed for pd.IndexSlice[]
  
  df1
  </code></pre>
  
  <p>Produces: (can't figure out how to format the output from Jupyter Notebook)</p>
  
  <pre><code>    AAPL    AMZN
      custom  price   custom  price
  Dates               
  2017-01-01  1   51  17  101
  2017-01-02  2   52  18  102
  2017-01-03  3   53  19  103
  2017-01-04  4   54  20  104
  2017-01-05  5   55  21  105
  2017-01-06  6   56  22  106
  </code></pre>
  
  <p><strong>Question:</strong>
  How can I create a 3rd column at the 2nd level of the MultiIndex that is the difference between <code>price</code> and <code>custom</code>? This should be calculated separately for each top column level, i.e. separately for AAPL and AMZN.</p>
  
  <p><strong>Attempted Solutions:</strong></p>
  
  <p>I tried using <code>pd.IndexSlice</code> in 2 ways, both give me all <code>NaNs</code>:</p>
  
  <pre><code>df1.loc[:,pd.IndexSlice[:,'price']].sub(df1.loc[:,pd.IndexSlice[:,'custom']])
  df1.loc[:,pd.IndexSlice[:,'price']] - df1.loc[:,pd.IndexSlice[:,'custom']]
  </code></pre>
  
  <p>Returns:</p>
  
  <pre><code>    AAPL    AMZN
      custom  price   custom  price
  Dates               
  2017-01-01  NaN NaN NaN NaN
  2017-01-02  NaN NaN NaN NaN
  2017-01-03  NaN NaN NaN NaN
  2017-01-04  NaN NaN NaN NaN
  2017-01-05  NaN NaN NaN NaN
  2017-01-06  NaN NaN NaN NaN
  </code></pre>
  
  <p>How can I add a third column with the difference?</p>
  
  <p>Thanks.</p>