Используйте SeriesGroupBy.value_counts
или size
с unstack
:
df = pd.DataFrame({
'A': [1000] * 10,
'B': [1000] * 2 + [1001] * 3 + [1002] * 5,
'C':[0,1] * 5
})
print (df)
A B C
0 1000 1000 0
1 1000 1000 1
2 1000 1001 0
3 1000 1001 1
4 1000 1001 0
5 1000 1002 1
6 1000 1002 0
7 1000 1002 1
8 1000 1002 0
9 1000 1002 1
df = df.groupby(['A','B'])['C'].value_counts().unstack(fill_value=0).reset_index()
#another solution
#df = pd.crosstab([df['A'], df['B']], df['C']).reset_index()
#solution 2
#df = df.groupby(['A','B','C']).size().unstack(fill_value=0).reset_index()
print (df)
C A B 0 1
0 1000 1000 1 1
1 1000 1001 2 1
2 1000 1002 2 3
А затем сложите и поделите:
df = df.rename(columns={0:'Count0',1:'Count1'})
df['Sum of Counts'] = df['Count0'] + df['Count1']
df['Count1/Sum of Counts'] = df['Count1'] / df['Sum of Counts']
print (df)
C A B Count0 Count1 Sum of Counts Count1/Sum of Counts
0 1000 1000 1 1 2 0.500000
1 1000 1001 2 1 3 0.333333
2 1000 1002 2 3 5 0.600000