如何在 pandas 中按组获取类别百分比

Question

Apologies if something similar has been asked before, I searched around but couldn't figure out a solution.

My dataset looks like such

data1 = {'Group':['Winner','Winner','Winner','Loser','Loser','Loser'],
        'MathStudy': ['Read','Read','Notes','Cheat','Cheat','Read'],
        'ScienceStudy': ['Notes','Read','Cheat','Cheat','Read','Notes']}
df1 = pd.DataFrame(data=data1)

I would like to get a % of total for each category for each group, as shown below. In my dataset the number of winners and losers changes, so a flexible solution is appreciated.

Thank you in advance!

Answer 1

Use DataFrame.melt with crosstab and normalize parameter:

df1 = df1.melt('Group', var_name='Type')

df2 = pd.crosstab([df1['Group'], df1['Type']], df1['value'], normalize=0)
print (df2)
value                   Cheat     Notes      Read
Group  Type                                      
Loser  MathStudy     0.666667  0.000000  0.333333
       ScienceStudy  0.333333  0.333333  0.333333
Winner MathStudy     0.000000  0.333333  0.666667
       ScienceStudy  0.333333  0.333333  0.333333
 

Last if need MultiIndex to columns with remove value column name add DataFrame.rename_axis with DataFrame.reset_index :

df2 = df2.rename_axis(columns=None).reset_index()
print (df2)
    Group          Type     Cheat     Notes      Read
0   Loser     MathStudy  0.666667  0.000000  0.333333
1   Loser  ScienceStudy  0.333333  0.333333  0.333333
2  Winner     MathStudy  0.000000  0.333333  0.666667
3  Winner  ScienceStudy  0.333333  0.333333  0.333333

Answer 2

@jezrael's solution is intuitive and what I would do first hand. However, I recently learned that melt usually performs poorly. Here's an alternative if performance is important, eg in codes that are used repeatedly:

g = df1.groupby('Group')
cols = ['MathStudy', 'ScienceStudy']
out = (pd.concat({col:g[col].value_counts(normalize=True) for col in cols})
   .unstack(level=-1, fill_value=0)
)

with run time:

2.9 ms ± 96.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Compared to melt approach:

9.44 ms ± 261 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Output:

                        Cheat     Notes      Read
MathStudy    Loser   0.666667  0.000000  0.333333
             Winner  0.000000  0.333333  0.666667
ScienceStudy Loser   0.333333  0.333333  0.333333
             Winner  0.333333  0.333333  0.333333

Note : pd.crosstab is essentially groupby() with some additional bookkeeping. And groupby on two columns are usually a lot slower.

Answer 3

Here is another alternative:

g = df.set_index('Group').stack().str.get_dummies().groupby(level=[0,1]).sum()
g.div(g.sum(axis=1),axis=0).round(2)