根据条件在groupby数据框上创建汇总表

Question

I have a pandas dataframe df that looks like

userid  trip_id segmentid   actual  prediction
  1       13       40          3       3
  1       6        2           1       1
  1       44       3           2       3
  2       70       19          1       1
  2       12       5           0       0

I need to create a summary dataframe dfsummary grouped on column userid , having three columns userid, correct_classified, incorrect_classified. If actual and prediction values are same then it is correct classified, otherwise incorrect classified.

I can count the correct_classfied on whole dataframe as

correct_classified = submission[(submission['Actual'] == submission['prediction'])]
incorrect_classified = submission[(submission['Actual'] != submission['prediction'])]

but don't getting an idea to create summary table grouped on user id, that should look like this

userid  correct_classified  incorrect_classified
  1             2                    1
  2             2                    0

Answer 1

You can use pd.crosstab after creating a conditional array:

flags = np.where(df['actual'].eq(df['prediction']), 'correct', 'incorrect')

res = pd.crosstab(df['userid'], flags)

print(res)

col_0   correct  incorrect
userid                    
1             2          1
2             2          0

Answer 2

You can also use pivot table ie

m = df['actual']==df['prediction']

# assign the conditions to new columns and aggregate.  
df.assign(correct_classified=m,incorrect_classified=~m).pivot_table(index='userid',
                                                                    aggfunc='sum',
                                                                    values=['correct_classified',
                                                                            'incorrect_classified'])

Output :

     correct_classified  incorrect_classified
userid                                          
1                      2.0                   1.0
2                      2.0                   0.0