Groupby and Named aggregation | Optimize dataframe generation in Pandas

Question

I have a dataframe in Pandas with some columns, something like this:

data = {
    'CODIGO_SINIESTRO': [10476434, 10476434, 4482524, 4482524, 4486110],
    'CONDICION': ['PASAJERO', 'CONDUCTOR', 'MOTOCICLISTA', 'CICLISTA', 'PEATON'],
    'EDAD': [62.0, 29.0, 26.0, 47.0, 33.0],
    'SEXO': ['MASCULINO', 'FEMENINO', 'FEMENINO', 'MASCULINO', 'FEMENINO']
}

df = pd.DataFrame(data)

Output:

    CODIGO_SINIESTRO    CONDICION       EDAD    SEXO
0   10476434            PASAJERO        62.0    MASCULINO
1   10476434            CONDUCTOR       29.0    MASCULINO
2   4482524             MOTOCICLISTA    26.0    MASCULINO
3   4482524             CICLISTA        47.0    MASCULINO
4   4486110             PEATON          33.0    FEMENINO

So, I want to create another dataframe grouped by 'CODIGO_SINIESTRO' column, and I want the following columns like result:

• 'CODIGO_SINIESTRO' : Id of the row.
• 'PROMEDIO_EDAD' : This column will store edad mean.
• 'CANTIDAD_HOMBRES' : This column will store masculine counts based on 'SEXO' column.
• 'CANTIDAD_HOMBRES' : This column will store femenine counts based on 'SEXO' column.

Finally I want five extra columns named equal to the four values possibles of 'CONDICION' column, this values will store 1 if value exist or 0 if not.

So, I wrote this solution and working as expect, however I have many rows in my dataset (150k+) and the solution is slow (5 minutes). This is my code:

df_final = df.groupby(['CODIGO_SINIESTRO']).agg(
    CANTIDAD_HOMBRES=pd.NamedAgg(column='SEXO', aggfunc=lambda x: (x=='MASCULINO').sum()),
    CANTIDAD_MUJERES=pd.NamedAgg(column='SEXO', aggfunc=lambda x: (x=='FEMENINO').sum()),
    PROMEDIO_EDAD=pd.NamedAgg(column='EDAD', aggfunc=np.mean),
    MOTOCICLISTA=pd.NamedAgg(column='CONDICION', aggfunc=lambda x: (x=='MOTOCICLISTA').any().astype(int)),
    CONDUCTOR=pd.NamedAgg(column='CONDICION', aggfunc=lambda x: (x=='CONDUCTOR').any().astype(int)),
    PEATON=pd.NamedAgg(column='CONDICION', aggfunc=lambda x: (x=='PEATON').any().astype(int)),
    CICLISTA=pd.NamedAgg(column='CONDICION', aggfunc=lambda x: (x=='CICLISTA').any().astype(int)),
    PASAJERO=pd.NamedAgg(column='CONDICION', aggfunc=lambda x: (x=='PASAJERO').any().astype(int))
).reset_index()

Output:

    CODIGO_SINIESTRO    CANTIDAD_HOMBRES    CANTIDAD_MUJERES    PROMEDIO_EDAD ...    
                                                    
 0    4482524                  1                      1               36.5  
 1    4486110                  0                      1               33.0  
 2    10476434                 1                      1               45.5


... MOTOCICLISTA    CONDUCTOR   PEATON  CICLISTA    PASAJERO
        1               0         0        1           0
        0               0         1        0           0
        0               1         0        0           1

How can I optimize this solution?, Are there other ways for resolving that?

Thank you.

Answer 1

Pre-aggregating with vectorized methods should be much more efficient (it turns out it was 100x faster):

df['PROMEDIO_EDAD']= df.groupby('CODIGO_SINIESTRO')['EDAD'].transform(np.mean)
df['CANTIDAD_HOMBRES'] = np.where(df['SEXO'] == 'MASCULINO', 1, 0)
df['CANTIDAD_MUJERES'] = np.where(df['SEXO'] == 'FEMENINO', 1, 0)
for col in df['CONDICION'].unique():
    df[col] = np.where(df['CONDICION'] == col, 1, 0)
df = df.groupby(['CODIGO_SINIESTRO', 'PROMEDIO_EDAD']).sum().reset_index().drop('EDAD', axis=1)
df.iloc[:,2:] = (df.iloc[:,2:] > 0).astype(int)
df
Out[1]: 
   CODIGO_SINIESTRO  PROMEDIO_EDAD  CANTIDAD_HOMBRES  CANTIDAD_MUJERES  \
0           4482524           36.5                 1                 1   
1           4486110           33.0                 0                 1   
2          10476434           45.5                 1                 1   

   PASAJERO  CONDUCTOR  MOTOCICLISTA  CICLISTA  PEATON  
0         0          0             1         1       0  
1         0          0             0         0       1  
2         1          1             0         0       0