percentage of sum in dataframe pandas

pandas, python

Solution

You can drop the `'None'` row like this:

df2 = df2.drop('None')

If you don't want it permanently dropped you don't have to assign that result back to `df2`.

Then you get your desired output with:

df2.apply(lambda c: c / c.sum() * 100, axis=0)
Out[11]: 
          Percentile1  Percentile2  Percentile3
value                                          
bottom      17.336683    22.251309    24.043716
top         17.336683    23.036649    22.404372
mediocre    65.326633    54.712042    53.551913

To just get straight to that result without permanently dropping the `None` row:

df2.drop('None').apply(lambda c: c / c.sum() * 100, axis=0)

Problem

i created the following dataframe by using pandas melt and groupby with value and variable. I used the following: df2 = pd.melt(df1).groupby(['value','variable'])['variable'].count().unstack('variable').fillna(0) ``` Percentile Percentile1 Percentile2 Percentile3 value None 0 16 32 48 bottom 0 69 85 88 top 0 69 88 82 mediocre 414 260 209 196 ``` I'm looking to create an output that excludes the 'None' row and creates a percentage of the sum of the 'bottom', 'top', and 'mediocre' rows. Desire output would be the following. ``` Percentile Percentile1 Percentile2 Percentile3 value bottom 0% 17.3% 22.3% 24.0% top 0% 17.3% 23.0% 22.4% mediocre 414% 65.3% 54.7% 53.6% ``` one of the main parts of this that i'm struggling with is creating a new row to equal an output. any help would be greatly appreciated!

Original source