title: New dataframe that is the result of value_counts on multiple columns in old dataframe
body: <p>I have a dataframe where the data looks like this.</p>
  <pre><code>        pack  pear  carrot  grape  berry  apple  sum 3rd pack  box
  0      1   1.0     3.0    1.0    2.0    1.0    8      NaN  1.0
  1      2   1.0     1.0    3.0    0.0    3.0    8      NaN  1.0
  2      3   1.0     1.0    3.0    1.0    3.0    9     True  1.0
  3      4   1.0     1.0    2.0    3.0    1.0    8      NaN  2.0
  4      5   2.0     3.0    4.0    0.0    0.0    9      NaN  2.0
  </code></pre>
  <p>The 5 columns with the fruit names only have values between 0 and 6 and I'd like to have a data frame that shows the distribution of those individual values. It should like exactly like this</p>
  <pre><code>         pear    carrot     grape     berry     apple
  0.0  0.163636  0.145455  0.090909  0.218182  0.236364
  1.0  0.472727  0.200000  0.218182  0.418182  0.454545
  2.0  0.218182  0.309091  0.290909  0.254545  0.145455
  3.0  0.090909  0.272727  0.236364  0.109091  0.127273
  4.0  0.036364  0.072727  0.090909       NaN  0.036364
  5.0  0.018182       NaN  0.036364       NaN       NaN
  6.0       NaN       NaN  0.036364       NaN       NaN 
  </code></pre>
  <p>I did this by creating a series for each fruit and then merging the series together</p>
  <pre><code>pears_per_pack = df['pear'].value_counts(normalize=True)
  carrots_per_pack = df['carrot'].value_counts(normalize=True)
  grapes_per_pack = df['grape'].value_counts(normalize=True)
  berrys_per_pack = df['berry'].value_counts(normalize=True)
  apples_per_pack = df['apple'].value_counts(normalize=True)
  
  df_lst = [pears_per_pack,carrots_per_pack,grapes_per_pack,berrys_per_pack,apples_per_pack]
  fruit_df = pd.concat(df_lst, axis = 1)
  </code></pre>
  <p>Now, this was only 5 columns, fairly easy to do by hand, but this whole thing is more or less a learning opportunity for me, and I figure this must break DRY protocol. SO I'm asking if there is a better way to do something like this that is more proper/ would be more applicable if it were a greater number of columns I needed to do this with.</p>